Ionic product of water and pH — calculate pH of 0.001M HCl and NaOH

easy CBSE JEE-MAIN NEET • NCERT Class 11 • 3 min read
Tags Equilibrium

Question

(a) Calculate the pH of 0.001 M HCl solution. (b) Calculate the pH of 0.001 M NaOH solution. (c) What is the ionic product of water at 25°C, and how does it change with temperature?

(NCERT Class 11, Chapter 7 — basic pH calculation, frequently tested)

Solution — Step by Step

HCl is a strong acid — it dissociates completely:

HCl→H++Cl−\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-

So [H+]=0.001=10−3[\text{H}^+] = 0.001 = 10^{-3} M.

pH=−log⁡[H+]=−log⁡(10−3)\text{pH} = -\log[\text{H}^+] = -\log(10^{-3}) pH=3\boxed{\text{pH} = 3}

NaOH is a strong base — complete dissociation:

NaOH→Na++OH−\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-

[OH−]=10−3[\text{OH}^-] = 10^{-3} M.

pOH=−log⁡[OH−]=−log⁡(10−3)=3\text{pOH} = -\log[\text{OH}^-] = -\log(10^{-3}) = 3

At 25°C: pH+pOH=14\text{pH} + \text{pOH} = 14

pH=14−3=11\boxed{\text{pH} = 14 - 3 = 11}

Water undergoes self-ionization:

H2O⇌H++OH−\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-

The ionic product: Kw=[H+][OH−]=10−14K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} at 25°C.

In pure water: [H+]=[OH−]=10−7[\text{H}^+] = [\text{OH}^-] = 10^{-7} M, so pH = 7.

KwK_w increases with temperature because the ionization of water is endothermic. At 60°C, Kw≈10−13K_w \approx 10^{-13}, and the pH of pure water drops below 7 — but the water is still neutral (equal [H+][\text{H}^+] and [OH−][\text{OH}^-]).

Why This Works

The pH scale is a logarithmic measure of hydrogen ion concentration. Each unit decrease in pH means a 10-fold increase in [H+][\text{H}^+]. This makes it convenient to express the enormous range of [H+][\text{H}^+] (from about 10010^{0} to 10−1410^{-14} M) on a simple 0-14 scale.

The relation pH + pOH = 14 holds only at 25°C. At other temperatures, use pH + pOH = pKwpK_w, where pKw=−log⁡KwpK_w = -\log K_w.

Alternative Method

For very dilute solutions (say, 10−810^{-8} M HCl), you cannot ignore the autoionization of water. The [H+][\text{H}^+] is not simply 10−810^{-8} but rather the solution of:

[H+]=c+Kw[H+][\text{H}^+] = c + \frac{K_w}{[\text{H}^+]}

For c=10−8c = 10^{-8}: [H+]≈1.05×10−7[\text{H}^+] \approx 1.05 \times 10^{-7}, giving pH ≈6.98\approx 6.98 (not 8!). This is a classic JEE trap.

For NEET, these pH problems are essentially free marks if you remember: (1) strong acids and bases dissociate completely, (2) pH = −log⁡[H+]-\log[\text{H}^+], (3) pH + pOH = 14 at 25°C. Practice with concentrations like 10−110^{-1}, 10−210^{-2}, 10−310^{-3} to build speed.

Common Mistake

A frequent error: students calculate the pH of 0.001 M NaOH as 3 (using −log⁡(0.001)-\log(0.001) directly). That gives the pOH, not the pH. For a base, always find pOH first, then convert: pH = 14 - pOH. If your pH for a basic solution comes out less than 7, something is wrong.

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