Question #35ef6

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Question #35ef6

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Answer 1 · 2017-03-25T18:37:27

You start from the buffer equation:

$p H = p K_{a} + {log}_{10} (\frac{[C_{3} H_{7} O_{2}^{-}]}{[C_{3} H_{7} C O_{2} H]})$ $pH=pK_a+log_(10)([[C_3H_7O_2^(-)]]/[[C_3H_7CO_2H]])$

Explanation:

And for your problem:

$5.17 = 4.069 + {log}_{10} (\frac{[C_{3} H_{7} O_{2}^{-}]}{[C_{3} H_{7} C O_{2} H]})$ $5.17=4.069+log_(10)([[C_3H_7O_2^(-)]]/[[C_3H_7CO_2H]])$

Clearly, ${log}_{10} (\frac{[C_{3} H_{7} O_{2}^{-}]}{[C_{3} H_{7} C O_{2} H]}) = 5.17 - 4.069 = 1.10$ $log_(10)([[C_3H_7O_2^(-)]]/[[C_3H_7CO_2H]])=5.17-4.069=1.10$

And we take antilogs.........

$10^{1.10} = \frac{[C_{3} H_{7} O_{2}^{-}]}{[C_{3} H_{7} C O_{2} H]} = 12.62$ $10^(1.10)=[[C_3H_7O_2^(-)]]/[[C_3H_7CO_2H]]=12.62$

But by spec. $[C_{3} H_{7} C O_{2} H] = 0.50 \cdot m o l \cdot L^{- 1}$ $[C_3H_7CO_2H]=0.50*mol*L^-1$ .

So $[C_{3} H_{7} O_{2}^{-}] = 6.31 \cdot m o l \cdot L^{- 1}$ $[C_3H_7O_2^-]=6.31*mol*L^-1$

For convenience, we would first prepare a $6.81 \cdot m o l \cdot L^{- 1}$ $6.81*mol*L^-1$ solution of $propionic acid$ $"propionic acid"$ (which as I recall would not have a very pleasant smell!), and then add a $6.31 \cdot m o l$ $6.31*mol$ $K O H$ $KOH$ to a $1 \cdot L$ $1*L$ volume.

Alternatively, we could add $6.31 \cdot m o l \times 96.07 \cdot g \cdot m o l^{- 1} = 606.1 \cdot g$ $6.31*molxx96.07*g*mol^-1=606.1*g$ $sodium propionate$ $"sodium propionate"$ to a $1 \cdot L$ $1*L$ volume of $0.500 \cdot m o l \cdot L^{- 1}$ $0.500*mol*L^-1$ $propionic acid$ $"propionic acid"$ . (The volumes would not change too much!)

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Question #35ef6

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