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Exercises · 6.48

Q.Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH
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Strong acids and bases dissociate completely in water, so their [H+][\text{H}^+] or [OH−][\text{OH}^-] equals the initial concentration. Use pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+] for acids and pOH=−log⁡[OH−]\text{pOH} = -\log[\text{OH}^-] then pH=14−pOH\text{pH} = 14 - \text{pOH} for bases.

Understanding pH and Complete Dissociation

The pH scale measures acidity through hydrogen ion concentration. When we say a substance undergoes complete dissociation, we mean every molecule splits into ions. HCl, HBr, NaOH, and KOH are all strong electrolytes—they dissociate 100% in dilute aqueous solution.

For a strong acid like HCl:

HCl→HX++ClX−\ce{HCl -> H^+ + Cl^-}

Every mole of HCl produces exactly one mole of HX+\ce{H^+}, so [HX+]=[HCl]initial[\ce{H^+}] = [\ce{HCl}]_{\text{initial}}.

For a strong base like NaOH:

NaOH→NaX++OHX−\ce{NaOH -> Na^+ + OH^-}

Similarly, [OHX−]=[NaOH]initial[\ce{OH^-}] = [\ce{NaOH}]_{\text{initial}}.

The key relationship connecting hydrogen and hydroxide ions in water at 25°C is:

Kw=[HX+][OHX−]=1.0×10−14K_w = [\ce{H^+}][\ce{OH^-}] = 1.0 \times 10^{-14}

This gives us pH+pOH=14\text{pH} + \text{pOH} = 14.

pH=−log⁡10[HX+]\text{pH} = -\log_{10}[\ce{H^+}]

pOH=−log⁡10[OHX−]\text{pOH} = -\log_{10}[\ce{OH^-}]

pH+pOH=14(at 25°C)\text{pH} + \text{pOH} = 14 \quad \text{(at 25°C)}

Solutions

(a) 0.003 M HCl

  1. Identify the species: HCl is a strong monoprotic acid.

  2. Find [HX+][\ce{H^+}]: Complete dissociation means [HX+]=0.003 M=3×10−3 M[\ce{H^+}] = 0.003 \text{ M} = 3 \times 10^{-3} \text{ M}.

  3. Calculate pH:

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

=−log⁡(3)−log⁡(10−3)= -\log(3) - \log(10^{-3})

=−0.477+3= -0.477 + 3

=2.52= 2.52

pH = 2.52

(b) 0.005 M NaOH

  1. Identify the species: NaOH is a strong base.

  2. Find [OHX−][\ce{OH^-}]: Complete dissociation gives [OHX−]=0.005 M=5×10−3 M[\ce{OH^-}] = 0.005 \text{ M} = 5 \times 10^{-3} \text{ M}.

  3. Calculate pOH:

pOH=−log⁡(5×10−3)\text{pOH} = -\log(5 \times 10^{-3})

=−log⁡(5)−log⁡(10−3)= -\log(5) - \log(10^{-3})

=−0.699+3= -0.699 + 3

=2.30= 2.30

  1. Convert to pH:

pH=14−pOH=14−2.30=11.70\text{pH} = 14 - \text{pOH} = 14 - 2.30 = 11.70

pH = 11.70

(c) 0.002 M HBr

  1. Identify the species: HBr is a strong monoprotic acid (one of the hydrohalic acids).

  2. Find [HX+][\ce{H^+}]: [HX+]=0.002 M=2×10−3 M[\ce{H^+}] = 0.002 \text{ M} = 2 \times 10^{-3} \text{ M}. …

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