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Chemistry Question on Equilibrium

What is [H+][H^+] in mol/L of a solution that is 0.20 M0.20 \,M in CH3COONaCH_3COONa and 0.10 M0.10 \,M in CH3COOHCH_3COOH ? (Ka  for  CH3COOH=1.8×10−5)(K_a\, \, for\, \, CH_3COOH = 1.8 \times 10^{-5})

A

3.5×10−43.5 \times 10^{-4}

B

1.1×10−51.1 \times 10^{-5}

C

1.8×10−51.8 \times 10^{-5}

D

9.0×10−69.0 \times 10^{-6}

Answer

9.0×10−69.0 \times 10^{-6}

Explanation

Solution

CH3COOHCH_3COOH (weak acid) and CH3COONaCH_3COONa (conjugated salt) form acidic buffer and for acidic buffer pH=pKa+log⁡[salt][acid]pH = pK_a + \log \frac{[salt]}{[acid]} and [H+]=−antilog pH[H^+] = -antilog\, pH
pH=−log⁡Ka+log⁡[salt][acid]pH = - \log K_a + \log \frac{[salt]}{[acid]}
[∴pKa=−log⁡Ka][\therefore pK_a = - \log K_a]
=−log⁡(1.8×10−5)+log⁡0.200.10= - \log (1.8 \times 10^{-5} ) + \log \frac {0.20}{0.10}
=4.74+log2= 4.74 + log 2
=4.74+0.3010=5.041= 4.74 + 0.3010 = 5.041
Now, [H+]=antilog(−5.045)[H^+ ] = antilog (-5.045)
=9.0×10−6mol/L= 9.0 \times 10^{-6} mol/L