What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? $K_{a}$ $K_a$ acetic acid = $1.8 x 10^{- 5}$ $1.8 x 10^-5$ ?

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What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? $K_{a}$ $K_a$ acetic acid = $1.8 x 10^{- 5}$ $1.8 x 10^-5$ ?

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Answer 1 · 2017-01-10T19:00:26

$\frac{[{AcO}^{-}]}{[HOAc]} = 1.76$ $(["AcO"^-])/(["HOAc"])=1.76$

Explanation:

$p H = p K_{a} - {log}_{10} (\frac{[{AcO}^{-}]}{[HOAc]})$ $pH=pK_a-log_10((["AcO"^-])/(["HOAc"]))$

And thus,

${log}_{10} (\frac{[{AcO}^{-}]}{[HOAc]}) = p K_{a} - p H = - {log}_{10} (1.8 \times 10^{- 5}) - 4.5 = 4.75 - 4.5 = 0.245$ $log_10((["AcO"^-])/(["HOAc"]))=pK_a-pH=-log_10(1.8xx10^-5)-4.5=4.75-4.5=0.245$

And thus $10^{0.245} = \frac{[A c O^{-}]}{[H O A c]} = 1.76$ $10^0.245=([AcO^-])/([HOAc])=1.76$

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What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? $K_{a}$ $K_a$ acetic acid = $1.8 x 10^{- 5}$ $1.8 x 10^-5$ ?

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Explanation:

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What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? K_aKaK_a acetic acid = 1.8 x 10^-51.8x10−51.8 x 10^-5?

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What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? $K_{a}$ $K_a$ acetic acid = $1.8 x 10^{- 5}$ $1.8 x 10^-5$ ?