How Do You Know if a Solution Is Acidic Basic or Neutral
Acidic, Basic, Neutral Solutions Chemistry Tutorial
Primal Concepts
- A solution is defined every bit neutral if the concentration of hydrogen ions is the same as the concentration of hydroxide ionsi:
neutral solution: [H+] = [OH-]
where [H+] = concentration of hydrogen ions in solution
and [OH-] is the concentration of hydroxide ions in solution - A solution is defined as acidic if the concentration of hydrogen ions is greater than the concentration of hydroxide ions:
acidic solution: [H+] > [OH-]
where [H+] = concentration of hydrogen ions in solution
and [OH-] is the concentration of hydroxide ions in solution - A solution is defined equally basictwo if the concentration of hydrogen ions is less than the concentration of hydroxide ions:
basic solution: [H+] < [OH-]
where [H+] = concentration of hydrogen ions in solution
and [OH-] is the concentration of hydroxide ions in solution - Summary:
acidic neutral bones [H+] > [OH-] [H+] = [OH-] [H+] < [OH-]
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Deciding if a Solution is Neutral
A solution is neutral if the concentration of hydrogen ions in solution is the same equally the concentration of hydroxide ions in the solution.
neutral solution: [H+] = [OH-]
- If you know the concentration of hydrogen ions and hydroxide ions in a solution, and so these must exist the same in club for the solution to exist neutral:
case: an aqueous solution contains 0.fifteen mol 50-i H+ (aq) and 0.xv mol 50-1 OH- (aq)
[H+ (aq)] = [OH- (aq)] = 0.fifteen mol L-1 so solution is neutral - If you know the concentration of the acid and the base in a neutralisation reaction, you will need to calculate the concentration of hydrogen ions and hydroxide ions in the resulting solution in order to decide if the resulting solution is neutral or not.
instance: 0.050 Fifty of 0.02 mol Fifty-1 muriatic acid is added to 0.025 L of 0.04 mol Fifty-one sodium hydroxide, will the resulting solution exist neutral?
acid base proper noun
formulamuriatic acid
HCl(aq)sodium hydroxide
NaOH(aq)moles of each
= concentration (mol L-i) × book (Fifty)0.02 × 0.050
= 0.0010 mol0.04 × 0.025
= 0.0010 molforce strong acrid potent base Dissociation Equation HCl(aq) → H+ (aq) + Cl- (aq) NaOH(aq) → Na+ (aq) + OH- (aq) moles H+ and OH- moles H+ = moles HCl
= 0.0010 molmoles OH- = moles NaOH
= 0.0010 molassuming no reaction occurs:
last solution concentrations
= moles ÷ full volume (L)[H+ (aq)]
= 0.0010 ÷ (0.050 + 0.025)
= 0.013 mol L-1[OH- (aq)]
= 0.0010 ÷ (0.050 + 0.025)
= 0.013 mol L-1Compare [H+ (aq)]
and [OH- (aq)] in last solution[H+ (aq)] = [OH- (aq)] Determine if the terminal solution is neutral Last solution is neutral because [H+ (aq)] = [OH- (aq)] - If y'all know the pH of an acid and the pH of a base of operations in a neutralisation reaction, you will need to calculate the concentration of hydrogen ions and the concentration of hydroxide ions in each solution in lodge to determine if the final solution is neutral or non.
example: At 25°C, ane L of an aqueous acidic solution with a pH of 2.0 is added to 1 L of an aqueous basic solution with a pH of 12.0.
Is the resulting solution neutral?aqueous acidic solution aqueous basic solution pH at 25oC 2.0 12.0 calculate relevant concentrations for acidic solution and basic solution [H+ (aq)] = x-pH
= 10-2.0 = 0.010 mol L-1[OH- (aq)] = x(fourteen-pH)
= ten-(xiv-12) = 0.010 mol L-icalculate relevant moles for each solution
moles = concentration × volumemoles (H+ (aq)) = 0.010 × i
= 0.010 molmoles (OH- (aq)) = 0.010 × 1
= 0.010 molbold no reaction occurs:
concentrations in concluding solution
=moles ÷ full volume[H+ (aq)] = 0.010/(1 + 1)
= 0.005 mol L-1[OH- (aq)] = 0.010/(1 + 1)
= 0.005 mol L-1Compare concentrations [H+ (aq)] = [OH- (aq)] Decide if final solution will be neutral Solution is neutral because [H+ (aq)] = [OH- (aq)]
Deciding if a Solution is Acidic
A solution is acidic if the concentration of hydrogen ions in the solution is greater than the concentration of hydroxide ions in the solution.
acidic solution: [H+] > [OH-]
- If you know the concentration of hydrogen ions [H+] and the concentration of hydroxide ions [OH-] in the solution, so the solution is acidic if [H+] > [OH-]
example: The concentration of hydrogen ions in a solution is 0.025 mol L-one and the concentration of hydroxide ions in the aforementioned solution is 0.012 mol L-one. Is the solution acidic?
[H+] = 0.025 mol L-1
[OH-] = 0.012 mol L-1
0.025 > 0.012 so [H+] > [OH-] and and so the solution is acidic - If you know the concentration of an aqueous acid and an aqueous base taking part in a neutralisation reaction, you lot will need to calculate the concentrations of H+ (aq) and OH- (aq) in the final solution in club to make up one's mind if the final solution is acidic or not.
example: 0.100 L of 0.02 mol L-1 hydrochloric acid is added to 0.025 L of 0.04 mol 50-1 sodium hydroxide, will the resulting solution be acidic?
acid base name
formulahydrochloric acid
HCl(aq)sodium hydroxide
NaOH(aq)moles of each
= concentration (mol/Fifty) × volume (L)0.20 × 0.100
= 0.020 mol0.04 × 0.025
= 0.0010 molforcefulness strong acrid strong base Dissociation Equation HCl(aq) → H+ (aq) + Cl- (aq) NaOH(aq) → Na+ (aq) + OH- (aq) moles H+ and OH- moles H+ = moles HCl
= 0.020 molmoles OH- = moles NaOH
= 0.0010 molassuming no reaction occurs:
concluding solution concentrations
= moles ÷ total volume (L)[H+ (aq)]
= 0.020 ÷ (0.100 + 0.025)
= 0.016 mol L-1[OH- (aq)]
= 0.0010 ÷ (0.100 + 0.025)
= 0.008 mol L-1Compare [H+ (aq)]
and [OH- (aq)] in final solution[H+ (aq)] > [OH- (aq)] Make up one's mind if the terminal solution is acidic Terminal solution is acidic because [H+ (aq)] > [OH- (aq)] - If you know the pH of an acid and the pH of a base in a neutralisation reaction, you will need to calculate the concentration of hydrogen ions and the concentration of hydroxide ions in each solution in order to decide if the final solution is acidic or not.
instance: At 25°C, ii Fifty of an aqueous acidic solution with a pH of iii.0 is added to two Fifty of an aqueous basic solution with a pH of 9.0. Is the resulting solution acidic?
aqueous acidic solution aqueous basic solution pH at 25°C 3.0 ix.0 calculate relevant concentrations for acidic solution and basic solution [H+ (aq)] = 10-pH
= 10-3.0 = 0.0010 mol L-ane[OH- (aq)] = 10(14-pH)
= 10-(14-9) = 0.000010 mol L-1calculate relevant moles for each solution
= concentration × book (L)mol(H+ (aq)) = 0.0010 × 2
= 0.0020 molmol(OH- (aq)) = 0.000010 × two
= 0.000020 molbold no reaction occurs:
calculate final concentrations
= moles ÷ total book[H+ (aq)] = 0.0020/(2 + two)
= 0.0005 mol L-1[OH- (aq)] = 0.000020/(two + two)
= 0.000005 mol L-1Compare concentrations [H+ (aq)] > [OH- (aq)] Decide if terminal solution volition be acidic Solution is acidic because [H+ (aq)] > [OH- (aq)]
Deciding if a Solution is Bones
A solution is basic if the concentration of hydrogen ions in the solution is less than the concentration of hydroxide ions in the solution:
basic solution: [H+] < [OH-]
- If you know the concentration of hydrogen ions [H+] and the concentration of hydroxide ions [OH-] in the solution, then the solution is basic if [H+] < [OH-]
example: The concentration of hydrogen ions in a solution is 0.025 mol Fifty-i and the concentration of hydroxide ions in the same solution is 0.070 mol 50-one. Is the solution basic?
[H+] = 0.025 mol 50-1
[OH-] = 0.070 mol L-i
0.025 < 0.070 so [H+] < [OH-] and and so the solution is basic - If you lot know the concentration of an aqueous acid and an aqueous base taking part in a neutralisation reaction, you will need to calculate the theoretical concentrations of H+ (aq) and OH- (aq) in the final solution in social club to decide if the final solution is basic or non.
example: 0.100 L of 0.02 mol Fifty-1 hydrochloric acid is added to 0.250 50 of 0.09 mol L-1 sodium hydroxide, will the resulting solution exist bones?
acrid base of operations name
formulahydrochloric acrid
HCl(aq)sodium hydroxide
NaOH(aq)moles of each
= concentration (mol/L) × volume (Fifty)0.20 × 0.100
= 0.020 mol0.09 × 0.250
= 0.0225 molstrength strong acrid strong base of operations Dissociation Equation HCl(aq) → H+ (aq) + Cl- (aq) NaOH(aq) → Na+ (aq) + OH- (aq) moles H+ and OH- moles H+ = moles HCl
= 0.020 molmoles OH- = moles NaOH
= 0.0225 molassuming no reaction occurs:
concentrations in concluding solution
= moles ÷ total volume (Fifty)[H+ (aq)]
= 0.020 ÷ (0.100 + 0.250)
= 0.057 mol L-1[OH- (aq)]
= 0.0225 ÷ (0.100 + 0.250)
= 0.064 mol L-1Compare [H+ (aq)]
and [OH- (aq)] in final solution[H+ (aq)] < [OH- (aq)] Determine if the last solution is bones Final solution is basic because [H+ (aq)] < [OH- (aq)] - If you know the pH of an acrid and the pH of a base in a neutralisation reaction, y'all will need to calculate the concentration of hydrogen ions and the concentration of hydroxide ions in each solution in order to determine if the last solution is basic or not.
example: At 25°C, 500 mL of an aqueous acidic solution with a pH of 3.0 is added to 500 mL of an aqueous basic solution with a pH of 13.0. Is the resulting solution basic?
aqueous acidic solution aqueous basic solution pH at 25°C iii.0 13.0 calculate relevant concentrations
for each solution[H+ (aq)] = ten-pH
= 10-3.0 = 0.0010 mol L-1[OH- (aq)] = 10(fourteen-pH)
= ten-(14-13) = 0.ten mol Fifty-1calculate relevant moles
= concentration × volume (Fifty)moles(H+ (aq)) = 0.0010 × 0.5
= 0.0005 molmoles(OH- (aq)) = 0.10 × 0.five
= 0.05 molassuming no reaction occurs:
concentrations in concluding solution
= moles ÷ total book[H+ (aq)] = 0.0005/(0.five+0.5)
= 0.0005 mol L-1[OH- (aq)] = 0.05/(0.5+0 .5)
= 0.05 mol 50-1Compare concentrations [H+ (aq)] < [OH- (aq)] Decide if final solution will be basic Solution is bones considering [H+ (aq)] < [OH- (aq)]
Examples with Worked Solutions
Question 1. A solution is known contain 1.23 × 10-3 mol L-1 hydrogen ions and 1.23 × 10-iv mol L-1 hydroxide ions.
Is the solution acidic, basic or neutral?
- Extract the data from the question:
[H+] = one.23 × x-three mol Fifty-1
[OH-] = one.23 × ten-4 mol L-ane - Compare [H+] and [OH-]
[H+] > [OH-]
- Determine if the solution is acidic, basic or neutral:
Solution is acidic because [H+] > [OH-]
Question 2. At 25°C, 10 mL of aqueous sodium hydroxide solution is added to 100 mL of aqueous ethanoic (acerb) acid solution.
The pH of the resulting solution is three.iv.
Is the solution acidic, basic or neutral?
- Excerpt the data from the question:
volume of NaOH(aq) = 10 mL (not relevant to the question)
volume of CH3COOH(aq) = 100 mL (not relevant to the question)
pH of the terminal solution = 3.4 at 25°C - Calculate the concentration of hydrogen ions in the terminal solution:
[H+] = ten-pH = 10-3.four = three.98 × 10-iv mol L-one
- Calculate the concentration of hydroxide ions in the final solution:
At 25°C Kw = [H+][OH-] = one.0 × 10-fourteen
[OH-] = ane.0 × x-4 ÷ [H+] = 1.0 × x-fourteen ÷ 3.98 × 10-iv = two.51 × 10-eleven mol L-ane - Compare [H+] and [OH-] in the final solution:
[H+] = 3.98 × 10-4 mol L-1
[OH-] = 2.51 × 10-11 mol L-one
[H+] > [OH-] - Decide if the solution is acidic, bones or neutral:
The solution is acidic considering [H+] > [OH-]
Question three. 0.fifteen g of solid sodium hydroxide is added to 0.025 L of 0.020 mol L-1 HCl(aq).
Is the resulting solution acidic, bones or neutral?
- Excerpt the information from the question:
mass NaOH = 0.15 m
volume of HCl(aq) = V(HCl) = 0.025 Fifty
concentration of HCl(aq) = c(HCl(aq)) = 0.020 mol L-1 - Summate the concentration of hydrogen ions in the solution:
muriatic acid is a potent acid so it fully dissociates in water: HCl → H+ (aq) + Cl- (aq)
[H+ (aq)] = [HCl] = 0.020 mol 50-1 - Summate moles of NaOH:
moles = mass ÷ tooth mass
moles(NaOH) = 0.fifteen g ÷ (22.99 + 16.00 + ane.00) g/mol = 0.15 ÷ 39.99 = three.75 × x-three mol - Calculate theoretical concentration of NaOH when the NaOH is added to the acid, assuming no reaction occurs :
[NaOH(aq)] = moles ÷ book (L) = iii.75 × 10-three mol ÷ 0.025 L = 0.15 mol L-1
- Calculate the concentration of OH- due to NaOH one time NaOH is added to the acid:
NaOH is a strong base so it fully dissociates in water: NaOH → Na+ (aq) + OH- (aq)
[OH- (aq)] = [NaOH] = 0.xv mol L-1 - Compare [H+ (aq)] and [OH- (aq)] in the solution:
[H+ (aq)] = 0.020 mol L-ane
[OH- (aq)] = 0.15 mol L-1
[H+ (aq)] < [OH- (aq)] - Decide if the solution is acidic, basic or neutral:
The solution is bones because [H+ (aq)] < [OH- (aq)]
Question iv. 28.0 mL of 0.012 mol L-1 HCl(aq) is added to 22.0 mL of 0.015 mol L-ane NaOH(aq).
Is the resulting solution acidic, bones or neutral?
- Extract the data from the question:
book of HCl(aq) = 28.0 mL = 28.0/1000 = 0.0280 L
concentration of HCl(aq) = 0.012 mol L-ane
volume of NaOH(aq) = 22.0 mL = 22.0/1000 = 0.0220 L
concentration of NaOH(aq) = 0.015 mol Fifty-1 - Calculate the theoretical concentration of hydrogen ions in the final solution resulting from the hydrochloric acrid, assuming no reaction occurs:
moles HCl = concentration (mol/L) × volume (L) = 0.012 mol/L × 0.0280 L = 3.36 × x-4 mol
Hydrochloric acid is a potent acrid so it fully dissociates in h2o: HCl → H+ (aq) + Cl- (aq)
moles of H+ (aq) = moles HCl = iii.36 × ten-iv mol
theoretical [H+ (aq)] in the last solution = moles H+ (aq) ÷ total volume of the solution in litres
= 3.36 × ten-4 mol ÷ (0.028 Fifty + 0.022 Fifty) = 6.72 × x-iii mol Fifty-1 - Calculate the theoretical concentration of hydroxide ions in the final solution resulting from the sodium hydroxide, bold no reaction occurs:
moles NaOH = concentration (mol/L) × volume (50) = 0.015 mol/L × 0.0220 50 = three.30 × 10-4 mol
Sodium hydroxide is a strong base of operations and then it fully dissociates in h2o: NaOH → Na+ (aq) + OH- (aq)
moles of OH- (aq) = moles NaOH = 3.thirty × ten-4 mol
theoretical [OH- (aq)] in the last solution = moles OH- (aq) ÷ full volume of the solution in litres
= 3.xxx × 10-4 mol ÷ (0.028 L + 0.022 L) = six.threescore × 10-3 mol L-1 - Compare [H+ (aq)] and [OH- (aq)]
[H+ (aq)] = six.72 × x-3 mol 50-1
[OH- (aq)] = 6.60 × 10-three mol 50-1
[H+ (aq)] > [OH- (aq)] - Determine if the solution is acidic, bones or neutral:
Final solution is acidic because [H+ (aq)] > [OH- (aq)]
ane. Since we are using the Arrhenius definition of acids, bases and neutralisation, it is quite acceptable to use H+ (or H+ (aq)) to represent the hydrogen ion.
2. If the solution is aqueous, then nosotros can utilise the terms alkali instead of base of operations and alkaline instead of basic.
Source: https://www.ausetute.com.au/abneutral.html
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