Calculate $p H$ $pH$ of buffer solution with $0.1$ $0.1$ $M$ $M$ each of $C H_{3} C O O H$ $CH_3COOH$ & $C H_{3} C O O N a$ $CH_3COONa$ . What will be the change in $p H$ $pH$ on addition of the following: ? ( $p K_{a}$ $pK_a$ = $1.8 \times 10^{- 5}$ $1.8 xx 10^-5$ ). The total volume of solution is $1$ $1$ $L$ $L$ .

Question

Calculate $p H$ $pH$ of buffer solution with $0.1$ $0.1$ $M$ $M$ each of $C H_{3} C O O H$ $CH_3COOH$ & $C H_{3} C O O N a$ $CH_3COONa$ . What will be the change in $p H$ $pH$ on addition of the following: ? ( $p K_{a}$ $pK_a$ = $1.8 \times 10^{- 5}$ $1.8 xx 10^-5$ ). The total volume of solution is $1$ $1$ $L$ $L$ .

(i) $0.01$ $0.01$ $M$ $M$ $H C l$ $HCl$
(ii) $0.01$ $0.01$ $M$ $M$ $N a O H$ $NaOH$

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Answer 1 · 2017-06-22T19:50:39

The buffer equation holds that.........

$p H = p K_{a} + {log}_{10} {\frac{[A^{-}]}{[H A]}}$ $pH=pK_a+log_10{([A^-])/([HA])}$

Explanation:

$p H = p K_{a} + {log}_{10} {\frac{[A^{-}]}{[H A]}}$ $pH=pK_a+log_10{([A^-])/([HA])}$

Where $H A$ $HA$ is the parent acid; and the $[A^{-}]$ $[A^-]$ is its conjugate. For the derivation of this formula, which an undergraduate should be able to perform, see here.

And thus for this problem..........

$p H = 4.76 + {log}_{10} {\frac{[A c O^{-}]}{[H O A c]}}$ $pH=4.76+log_10{([AcO^-])/([HOAc])}$

And note that $4.76 = - {log}_{10} (1.8 \times 10^{- 5})$ $4.76=-log_10(1.8xx10^-5)$

Now the starting conditions specify that there are equal concentrations of acetate and free acid, and thus $p H = p K_{a} = 4.76$ $pH=pK_a=4.76$ . Why? Because ${log}_{10} 1 = 0$ $log_(10)1=0$ , so given the equation, $[A c O^{-}] = [H O A c]$ $[AcO^-]=[HOAc]$ , and $\frac{[A c O^{-}]}{[H O A c]} = 1$ $([AcO^-])/([HOAc])=1$ .

Anyway, work thru the problems, and post them back in this thread. The $p H$ $pH$ will be round about $4.76$ $4.76$ , i.e. the $p K_{a}$ $pK_a$ of the parent acid.

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(i) $0.01$ $0.01$ $M$ $M$ $H C l$ $HCl$
(ii) $0.01$ $0.01$ $M$ $M$ $N a O H$ $NaOH$

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Explanation:

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Calculate pHpH of buffer solution with 0.10.1 MM each of CH3COOHCH_3COOH & CH3COONaCH_3COONa. What will be the change in pHpH on addition of the following: ? (pKapK_a = 1.8×10−51.8 xx 10^-5). The total volume of solution is 11 LL.

(i) 0.010.01 MM HClHCl (ii) 0.010.01 MM NaOHNaOH

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Explanation:

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(i) $0.01$ $0.01$ $M$ $M$ $H C l$ $HCl$
(ii) $0.01$ $0.01$ $M$ $M$ $N a O H$ $NaOH$