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Prove the buffer action of acetic acid and sodium acetate by the addition of 0.01 mol of solid sodium hydroxide. |
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Answer» Solution :(i) Consider ONE litre to buffer solution containing 0.8 cm `CH_3COOH` and 0.8 cm `CH_3COONa`. Assume that the VOLUME changes due to the addition of 0.01 mol of solid NaOH is negligible. `K_a` for `CH_3COOH` is `1.8 times 10^-5` (ii) `Ch_3COOH(AQ) leftrightarrow^(H_2O(l)) CH_3COO^(-) (aq) +H_(aq)^+` 0.8-aq a a `CH_3COONa_(aq) to^(H_2O) CH_3COO^(-) (aq)+Na_(aq)^+` 0.8 0.8 0.8 (iii) The dissociation constant for `Ch_3COOH` is given by `K_a=([CH_3COO^-][H^+])/([CH_3COOH])` `[H^+]=(K_a[CH_3COOH])/([CH_3COO^-])` The above expression shows that the concentration of `H^+` is directly proportional to `([CH_3COOH])/([CH_3COO^-])` degree of dissociation of `CH_3COOH-a` (iv) `[CH_3COOH]=0.8-a and [CH_3COO^-]=a+0.8` `therefore [H^+]=K_a.([0.8-a])/([0.8+a])` `If a lt lt0.8,0.8-a approx0.8` `0.8+a approx0.8` `therefore [H^+]=(K_a.[0.8])/([0.8])=K_a` `[H^+]=K_a` (v) Given that `K_a` for `CH_3COOH` is `1.8 times 10^-5` `[H^+]=1.8 times10^-5` `pH=-log[H^+]` `=-log[1.8 times10^-5]` `=5-log 1.8` `=5-0.26` `pH=4.74` (vi) After adding 0.1 mol NaOH to 1 litre of buffer. Given that volume change due to the addition of NaOH is negligible `therefore [OH^-]=0.01M` The consumption of `OH^-` are represented by the following EQUATION `CH_3COOH_(aq) leftrightarrowCH_3COO_(aq)^(-) +H^+ (aq)` 0.8-aa a `CH_3COONa_(aq) toCH_3COO^(-)+Na^(+) (aq)` 0.8 0.8 0.8 `therefore[CH_3COOH]=0.8-a-0.01=0.79-a` `[CH_3COO^-]=a+0.8+0.01` `=0.81+aa lt lt 0.8` `0.79-a approx0.79 and 0.81+a approx 0.81` `therefore [H^+]=1.8 times 10^-5 times 0.79/0.81` `[H^+]=1.76 times10^-5` `therefore pH=-log(1.76 times10^-5)` `=5- log 1.76` `=5-0.25` `pH=4.75` (vii) The addition of a strong base (0.01M NaOH) increased the pH only slightly i.e., from 4.74to 4.75. SO the buffer action is verified. |
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