JEE MAIN - Chemistry (2021 - 18th March Morning Shift - No. 16)
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ___________ M. (Round off to the Nearest Integer). [Given : pKa (acetic acid) = 4.74]
Answer
10
Explanation
$$C{H_3}COOH + C{H_3}COONa$$ [Acidic Buffer]
$${pH} = {p{ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$$
$$ \Rightarrow 5.74 = 4.74 + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$$
$$ \Rightarrow 1 = \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$$
$$ \Rightarrow {{[C{H_3}COONa]} \over {[C{H_3}COOH]}} = 10$$
$$ \Rightarrow [C{H_3}COONa] = 10 \times 1 = 10$$
$${pH} = {p{ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$$
$$ \Rightarrow 5.74 = 4.74 + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$$
$$ \Rightarrow 1 = \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$$
$$ \Rightarrow {{[C{H_3}COONa]} \over {[C{H_3}COOH]}} = 10$$
$$ \Rightarrow [C{H_3}COONa] = 10 \times 1 = 10$$
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