Chemistry Clarity · chemistryclarity.com

Acid–Base Properties of Salts

Why are some salt solutions acidic or basic, and how do you calculate their pH?

IntermediateAcids & BasesLast reviewed 6 October 2026
Download lesson (PDF)

What is it?

A salt is the ionic compound formed when an acid neutralizes a base. You might expect every salt solution to be neutral, but many are not:

  • Sodium chloride solution has pH 7.
  • Sodium ethanoate (sodium acetate) solution is basic.
  • Ammonium chloride solution is acidic.

The reason is that some ions react with water. This reaction is called hydrolysis.

  • The conjugate base of a weak acid (such as CHX3COOX−\ce{CH3COO-} or FX−\ce{F-}) takes a proton from water, making OHX−\ce{OH-}:
CHX3COOX−(aq)+HX2O(l)⇌CHX3COOH(aq)+OHX−(aq)\small\begin{aligned} &\ce{CH3COO-(aq) + H2O(l)} \\[4pt] &\quad \ce{<=> CH3COOH(aq) + OH-(aq)} \end{aligned}
  • The conjugate acid of a weak base (such as NHX4X+\ce{NH4+}) gives a proton to water, making HX3OX+\ce{H3O+}:
NHX4X+(aq)+HX2O(l)⇌NHX3(aq)+HX3OX+(aq)\small\begin{aligned} &\ce{NH4+(aq) + H2O(l)} \\[4pt] &\quad \ce{<=> NH3(aq) + H3O+(aq)} \end{aligned}
  • The conjugates of strong acids and bases (ClX−\ce{Cl-}, NOX3X−\ce{NO3-}, NaX+\ce{Na+}, KX+\ce{K+}) are far too weak to react with water: they are neutral.

Key idea

Look at where the salt came from. The “parent” that is stronger wins. Strong acid + strong base gives a neutral salt; strong acid + weak base gives an acidic salt; weak acid + strong base gives a basic salt.

Why does it matter?

  • Titration end points. At the equivalence point of a weak acid–strong base titration, the solution contains only the salt, so the pH is above 7. This decides which indicator to use.
  • Everyday chemistry. Baking soda and washing soda solutions are basic; ammonium salts in fertilizers make soil more acidic.
  • Medicine and biology. Many drugs are given as salts, and their pH in solution affects how they are absorbed.

How does it work?

1. Predicting acidic, basic or neutral

Salt made fromExampleIon that hydrolysesSolution
Strong acid + strong baseNaCl, KNO₃noneneutral, pH 7
Strong acid + weak baseNH₄ClNHX4X+\ce{NH4+}acidic, pH below 7
Weak acid + strong baseCH₃COONa, NaFCHX3COOX−\ce{CH3COO-}, FX−\ce{F-}basic, pH above 7

(For a salt of a weak acid and a weak base, compare KaK_\text{a} of the cation with KbK_\text{b} of the anion.)

2. Ka × Kb = Kw

For any conjugate acid–base pair in water:

Ka×Kb=Kw=1.0×10−14 at 25 °C\small\begin{aligned} &K_\text{a} \times K_\text{b} = K_\text{w} = 1.0 \\[4pt] &\quad \times 10^{-14}\ \text{at 25 °C} \end{aligned}

So the weaker the acid, the stronger its conjugate base. You can find KbK_\text{b} for CHX3COOX−\ce{CH3COO-} from KaK_\text{a} for ethanoic acid, and KaK_\text{a} for NHX4X+\ce{NH4+} from KbK_\text{b} for ammonia.

3. Calculating the pH of a salt solution

Treat the hydrolysing ion as a weak base (or weak acid) and use the same method as for weak acids:

  1. Find KbK_\text{b} (or KaK_\text{a}) of the ion from KaKb=KwK_\text{a} K_\text{b} = K_\text{w}.
  2. Set up the equilibrium: x=[OHX−]x = [\ce{OH-}] (or [HX3OX+][\ce{H3O+}]), and x≈KCx \approx \sqrt{K C} if xx is small.
  3. Convert to pH.

Think of it like this

Think of a tug of war between two teams, the parent acid and the parent base. If both teams are strong, they cancel and the rope doesn’t move (neutral). If one is strong and the other weak, the strong side’s partner (the conjugate ion of the weak side) still pulls a little: the solution leans towards the weak parent’s opposite.

More precisely

Small, highly charged metal ions such as AlX3+\ce{Al^3+} and FeX3+\ce{Fe^3+} also make solutions acidic: the hydrated ion, e.g. [Al(HX2O)X6]X3+\ce{[Al(H2O)6]^3+}, releases a proton from one of its water molecules. The relationship KaKb=KwK_\text{a}K_\text{b} = K_\text{w} can also be written pKa+pKb=14.00\text{p}K_\text{a} + \text{p}K_\text{b} = 14.00 at 25 °C. KwK_\text{w} increases with temperature, so neutral pH is slightly below 7 in hot water.

Visualise it

Summary chart. Acid plus base gives a salt X plus Y minus and water. Look at each ion. Cations can only make a solution acidic: (1) a cation from a weak base, such as ammonium, gives H3O+ (acidic); (1) a small, highly charged metal ion such as Al3+ or Fe3+ also gives H3O+ from its attached water (acidic); (2) a cation from a strong base, such as Na+ or K+, does not react (neutral). Anions can only make a solution basic: (3) an anion from a weak acid, such as ethanoate, fluoride or carbonate, gives OH- (basic); (4) an anion from a strong acid, such as chloride or nitrate, does not react (neutral). Combined effect: 1 plus 4 is acidic, for example NH4Cl and AlCl3; 2 plus 3 is basic, for example CH3COONa and Na2CO3; 2 plus 4 is neutral, for example NaCl and KNO3; 1 plus 3 depends on comparing Ka of the cation with Kb of the anion, for example NH4F is acidic.
Summary: judge each ion of the salt separately, then combine their effects.

Applying the chart to four solutions gives these pH values:

A chart of salts on a pH scale. Ammonium chloride, from a strong acid and a weak base, is on the acidic side. Sodium chloride, from a strong acid and a strong base, is at pH 7. Sodium ethanoate and sodium fluoride, from weak acids and a strong base, are on the basic side.
The parent acid and base decide whether a salt solution is acidic, neutral or basic.

Worked example

Worked example: Acidic, basic or neutral?

Question: Predict whether solutions of (a) KNO₃ (b) NH₄NO₃ (c) Na₂CO₃ (d) KF are acidic, basic or neutral.

  1. (a) From HNOX3\ce{HNO3} (strong) and KOH\ce{KOH} (strong): neutral.
  2. (b) NHX4X+\ce{NH4+} comes from NHX3\ce{NH3} (weak base); NOX3X−\ce{NO3-} is neutral: acidic.
  3. (c) COX3X2−\ce{CO3^2-} comes from a weak acid; NaX+\ce{Na+} is neutral: basic.
  4. (d) FX−\ce{F-} comes from HF (weak acid): basic.

Worked example: Finding Kb from Ka

Question: Find KbK_\text{b} for the ethanoate ion, given Ka(CHX3COOH)=1.8×10−5K_\text{a}(\ce{CH3COOH}) = 1.8 \times 10^{-5}.

Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10\small\begin{aligned} &K_\text{b} = \frac{K_\text{w}}{K_\text{a}} \\[4pt] &= \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \\[4pt] &= 5.6 \times 10^{-10} \end{aligned}

Worked example: pH of a basic salt

Question: Calculate the pH of 0.10 mol/L sodium ethanoate.

  1. CHX3COOX−\ce{CH3COO-} is a weak base with Kb=5.6×10−10K_\text{b} = 5.6 \times 10^{-10}. Let x=[OHX−]x = [\ce{OH-}]:

    x≈KbC=5.6×10−10×0.10=7.48×10−6 mol/L\begin{aligned} &x \approx \sqrt{K_\text{b} C} \\[4pt] &= \sqrt{5.6 \times 10^{-10} \times 0.10} \\[4pt] &= 7.48 \times 10^{-6}\ \text{mol/L} \end{aligned}

  2. Check: xx is far below 5 % of 0.10 mol/L, so the approximation holds.

  3. pOH and pH:

    pOH=−log⁡(7.48×10−6)=5.13pH=14.00−5.13=8.87\small\begin{aligned} &\text{pOH} = -\log(7.48 \times 10^{-6}) \\[4pt] &= 5.13 \\[4pt] &\text{pH} = 14.00 - 5.13 = 8.87 \end{aligned}

Worked example: pH of an acidic salt

Question: Calculate the pH of 0.20 mol/L ammonium chloride. (Kb(NHX3)=1.8×10−5K_\text{b}(\ce{NH3}) = 1.8 \times 10^{-5})

  1. Ka(NHX4X+)=1.0×10−141.8×10−5=5.6×10−10K_\text{a}(\ce{NH4+}) = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}

  2. Let x=[HX3OX+]x = [\ce{H3O+}]:

    x≈5.6×10−10×0.20=1.058×10−5 mol/L\begin{aligned} &x \approx \sqrt{5.6 \times 10^{-10} \times 0.20} \\[4pt] &= 1.058 \times 10^{-5}\ \text{mol/L} \end{aligned}

  3. pH=−log⁡(1.058×10−5)=4.98\text{pH} = -\log(1.058 \times 10^{-5}) = 4.98

Common mistake

Common mistake: Assuming every salt is neutral

“Salt” does not mean “neutral”. Only salts of a strong acid and a strong base give pH 7. Check the parents of both ions.

Common mistake: Using Ka of the parent acid directly

The base in a sodium ethanoate solution is CHX3COOX−\ce{CH3COO-}, not ethanoic acid. Convert first: Kb=Kw/KaK_\text{b} = K_\text{w}/K_\text{a}.

Common mistake: Forgetting to convert pOH to pH

For a basic salt, the calculation gives [OHX−][\ce{OH-}] and pOH. Finish with pH = 14.00 − pOH.

Notation note

  • Hydrolysis is shown with equilibrium arrows (⇌) because only a small fraction of the ions react.
  • Units: KaK_\text{a} and KbK_\text{b} are written without units (they are ratios of concentrations relative to 1 mol/L).

Remember this

Remember this

  • Strong acid + strong base → neutral salt; strong acid + weak base → acidic; weak acid + strong base → basic.
  • Conjugates of strong acids and bases (ClX−\ce{Cl-}, NOX3X−\ce{NO3-}, NaX+\ce{Na+}, KX+\ce{K+}) don’t hydrolyse.
  • Ka×Kb=Kw=1.0×10−14K_\text{a} \times K_\text{b} = K_\text{w} = 1.0 \times 10^{-14} at 25 °C.
  • pH of a salt: find K of the hydrolysing ion, then x≈KCx \approx \sqrt{KC}, then pH.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What is hydrolysis of a salt?
    Answer

    The reaction of an ion with water, producing H₃O⁺ or OH⁻ and changing the pH.

  2. Question
    Is a salt of a strong acid and a strong base acidic, basic or neutral?
    Answer

    Neutral (pH 7), e.g. NaCl, KNO₃.

  3. Question
    Is a salt of a strong acid and a weak base acidic, basic or neutral?
    Answer

    Acidic, e.g. NH₄Cl: NH₄⁺ gives H⁺ to water.

  4. Question
    Is a salt of a weak acid and a strong base acidic, basic or neutral?
    Answer

    Basic, e.g. CH₃COONa: CH₃COO⁻ takes H⁺ from water, forming OH⁻.

  5. Question
    Which ions do not hydrolyse?
    Answer

    Conjugates of strong acids and bases: Cl⁻, Br⁻, I⁻, NO₃⁻, Na⁺, K⁺ (and other Group 1 ions).

  6. Question
    State the relationship between Ka and Kb for a conjugate pair.
    Answer

    Ka×Kb=Kw=1.0×10−14K_\text{a} \times K_\text{b} = K_\text{w} = 1.0 \times 10^{-14} at 25 °C

  7. Question
    Ka of ethanoic acid is 1.8 × 10⁻⁵. What is Kb of the ethanoate ion?
    Answer

    1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰

  8. Question
    Why is the pH at the equivalence point of a weak acid–strong base titration above 7?
    Answer

    The solution contains the salt of the weak acid, whose anion is a weak base.

  9. Question
    Why are solutions of Al³⁺ or Fe³⁺ salts acidic?
    Answer

    The small, highly charged hydrated ions release H⁺ from their attached water molecules.

  10. Question
    Outline how to calculate the pH of 0.10 mol/L sodium ethanoate.
    Answer

    Kb = Kw/Ka → [OH⁻] ≈ √(Kb × C) → pOH → pH = 14.00 − pOH (answer 8.87).

Quiz

7 questions

  1. Question 1EasyWhich salt gives a basic solution?
    Show answer

    Answer: K₂CO₃

    CO₃²⁻ is the conjugate base of a weak acid, so it takes protons from water and forms OH⁻. K⁺ is neutral.

  2. Question 2EasyWhich salt gives an acidic solution?
    Show answer

    Answer: NH₄Cl

    NH₄⁺ is the conjugate acid of the weak base NH₃ and donates a proton to water. Cl⁻ is neutral.

  3. Question 3EasyWhy is a solution of NaCl neutral?
    Show answer

    Answer: Neither Na⁺ nor Cl⁻ reacts with water to change [H⁺] or [OH⁻]

    Both ions come from a strong base and a strong acid, so they are far too weak to react with water.

  4. Question 4MediumKa of HCN is 4.9 × 10⁻¹⁰. What is Kb of CN⁻?
    Show answer

    Answer: 2.0 × 10⁻⁵

    Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 4.9 × 10⁻¹⁰ = 2.0 × 10⁻⁵. A very weak acid has a fairly strong conjugate base.

  5. Question 5HardWhat is the pH of 0.10 mol/L NaCN? (Kb of CN⁻ = 2.0 × 10⁻⁵)
    Show answer

    Answer: 11.15

    [OH⁻] ≈ √(2.0 × 10⁻⁵ × 0.10) = 1.41 × 10⁻³ mol/L, pOH = 2.85, pH = 14.00 − 2.85 = 11.15. 2.85 is the pOH.

  6. Question 6MediumThe equivalence point of a titration of ethanoic acid with NaOH has a pH of about:
    Show answer

    Answer: 9

    At equivalence the solution contains sodium ethanoate, a basic salt, so the pH is above 7 (about 8.7 for 0.050 mol/L).

  7. Question 7HardPut these 0.10 mol/L solutions in order of increasing pH: NaCN, NH₄Cl, NaCl, CH₃COONa.
    Show answer

    Answer: NH₄Cl, NaCl, CH₃COONa, NaCN

    NH₄Cl is acidic; NaCl is neutral; CH₃COO⁻ (Kb 5.6 × 10⁻¹⁰) is a weaker base than CN⁻ (Kb 2.0 × 10⁻⁵), so NaCN has the highest pH.

Notes and downloads

  • Worksheet

    Acid–Base Properties of Salts Worksheet

    8 questions on predicting acidic, basic and neutral salts, Ka × Kb = Kw, and calculating the pH of salt solutions and titration equivalence points. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/acid-base-properties-of-salts/

Spotted a mistake? Let us know and we'll fix it.