JEE Challenger
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ChemMedAdv 2022 • Paper 1

Calculation of pH of Carbonic Acid Salt Solution Mix

A solution is prepared by mixing 0.01 mol0.01\text{ mol} each of H2CO3\text{H}_2\text{CO}_3, NaHCO3\text{NaHCO}_3, Na2CO3\text{Na}_2\text{CO}_3, and NaOH\text{NaOH} in 100 mL100\text{ mL} of water. pH\text{pH} of the resulting solution is ______.

[Given: pKa1pK_{a1} and pKa2pK_{a2} of H2CO3\text{H}_2\text{CO}_3 are 6.376.37 and 10.3210.32, respectively; log⁡2=0.30\log 2 = 0.30]

Official Numerical Answer10.02

Topics & Concepts

EquilibriumAcid-Base EquilibriumMole Concept

Step-by-Step Solution

To determine the pH\text{pH} of the resulting solution, we need to analyze the chemical reactions that take place upon mixing the components.

Step 1: Identify the species and their initial mole amounts

The solution contains:

  • H2CO3=0.01 mol\text{H}_2\text{CO}_3 = 0.01\text{ mol}
  • NaHCO3\text{NaHCO}_3 (which dissociates to yield HCO3−\text{HCO}_3^-) =0.01 mol= 0.01\text{ mol}
  • Na2CO3\text{Na}_2\text{CO}_3 (which dissociates to yield CO32−\text{CO}_3^{2-}) =0.01 mol= 0.01\text{ mol}
  • NaOH\text{NaOH} (which dissociates completely to yield OH−\text{OH}^-) =0.01 mol= 0.01\text{ mol}

Step 2: Neutralization reaction

NaOH\text{NaOH} is a strong base and will react with the strongest acid present in the solution, which is carbonic acid (H2CO3\text{H}_2\text{CO}_3).

The neutralization reaction is: H2CO3(aq)+OH−(aq)→HCO3−(aq)+H2O(l)\text{H}_2\text{CO}_3\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{HCO}_3^-\text{(aq)} + \text{H}_2\text{O(l)}

Since 0.01 mol0.01\text{ mol} of NaOH\text{NaOH} reacts completely with 0.01 mol0.01\text{ mol} of H2CO3\text{H}_2\text{CO}_3:

  • Moles of H2CO3 remaining=0.01−0.01=0 mol\text{Moles of } \text{H}_2\text{CO}_3 \text{ remaining} = 0.01 - 0.01 = 0\text{ mol}
  • Moles of OH− remaining=0.01−0.01=0 mol\text{Moles of } \text{OH}^- \text{ remaining} = 0.01 - 0.01 = 0\text{ mol}
  • Moles of HCO3− produced=0.01 mol\text{Moles of } \text{HCO}_3^- \text{ produced} = 0.01\text{ mol}

Step 3: Total moles in the final solution

After the neutralization reaction, the amounts of species present in the total volume (V=100 mLV = 100\text{ mL}) are:

  • Total moles of HCO3−=0.01 mol (initial)+0.01 mol (formed)=0.02 mol\text{Total moles of } \text{HCO}_3^- = 0.01\text{ mol (initial)} + 0.01\text{ mol (formed)} = 0.02\text{ mol}
  • Total moles of CO32−=0.01 mol\text{Total moles of } \text{CO}_3^{2-} = 0.01\text{ mol}

Step 4: Calculate the pH of the buffer solution

The mixture of HCO3−\text{HCO}_3^- (weak acid) and CO32−\text{CO}_3^{2-} (conjugate base) forms a buffer system described by the equilibrium: HCO3−(aq)⇌H+(aq)+CO32−(aq)\text{HCO}_3^-\text{(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CO}_3^{2-}\text{(aq)}

The equilibrium constant for this reaction is Ka2K_{a2}. Applying the Henderson-Hasselbalch equation for acid buffer: pH=pKa2+log⁡10([CO32−][HCO3−])\text{pH} = pK_{a2} + \log_{10}\left( \frac{[\text{CO}_3^{2-}]}{[\text{HCO}_3^-]} \right)

Substitute the mole amounts into the equation (since the volume terms cancel out): pH=pKa2+log⁡10(n(CO32−)n(HCO3−))\text{pH} = pK_{a2} + \log_{10}\left( \frac{n(\text{CO}_3^{2-})}{n(\text{HCO}_3^-)} \right)

Given pKa2=10.32pK_{a2} = 10.32 and log⁡102=0.30\log_{10} 2 = 0.30: pH=10.32+log⁡10(0.010.02)\text{pH} = 10.32 + \log_{10}\left( \frac{0.01}{0.02} \right) pH=10.32+log⁡10(12)\text{pH} = 10.32 + \log_{10}\left( \frac{1}{2} \right) pH=10.32−log⁡102\text{pH} = 10.32 - \log_{10} 2 pH=10.32−0.30=10.02\text{pH} = 10.32 - 0.30 = 10.02

Thus, the pH\text{pH} of the resulting solution is 10.0210.02.

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