Question

✪ Chemistry ✪
What is PoH of a weak acid having concentration $\sf 10^{-4}$ and % dissociation is 4%.

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Answer 1

The chemical equation for dissociation of weak acid HA is represented as follows : $\sf{H A(a q)+H_{2}O \rightarrow H_{3} O^{+}(a q)+A^{-}(a q)}$

Calculating the concentration of acid using percent dissociation and initial concentration of acid as follows :

$\sf{\textsf{ Percent dissociation }=\dfrac{[H A]_{\textsf{dissociated }}}{[H A]_{\textsf{initial }}} \times 100}$

$\to\sf{\textsf{ Percent dissociation }\times [H A]_{\textsf{initial }}=[H A]_{\textsf{dissociated }}\times 100}$

$\to\sf{[H A]_{\textsf{dissociated }}=\dfrac{\textsf{ Percent dissociation } \times[H A]_{\textsf{initial }}}{100}}$

$\to\sf{[H A]_{\textsf{dissociated }}=\dfrac{4 \times 10^{-4}}{100}}$

$\to\sf{[H A]_{\textsf{dissociated }}=0.000004\:M}$

Thus, the concentration of hydronium ion will be equal to the concentration of dissociated acid that is 0.000004 M.

Now, the pH of acid will be calculated as follows :

$\sf{p H=-\log \left[H^{+}\right]}$

$\to\sf{p H=-\log \left[0.000004\right]}$

$\to\sf{p H=5.4}$

Now, the pOH of the acid will be calculated as follows :

$\sf{p H+p O H=14}$

$\to\sf{p H+p O H-p H=14-pH}$

$\to\sf{pOH=14-pH}$

$\to\sf{pOH=14-5.4}$

$\boxed{\boxed{\to\sf{pOH=8.6}}}$

Therefore, the pOH of weak acid is 8.6

Answer 2

Explanation:

✪ Chemistry ✪

What is PoH of a weak acid having concentration $\sf 10^{-4}$ and % dissociation is 4%.

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