Question

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$\sf{calculate \: the \: pH\: of \: a \: {10}^{ - 3}M \: solution \: of \: Ca(OH)\tiny{2}}\sf{{\: if \: it \: undergoes \: complete \: ionisation}}$
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Answer 1

Answer:

Step-by-step explanation:

To Find :-

pH of 10^(-3)M solution of Ca(OH)_2.

Solution :-

We know that :-

$pH=-\log_{10}{[H^+]}$

$pOH=-\log_{10}{[OH^-]}$

According to Question :-

$pOH=-\log_{10}{10^{-3}}$

$pOH=-(-3)\log_{10}{10}$

$pOH=3\log_{10}{10}$

$pOH=3$

$\therefore \sf\log_{a}{a}=1$

We know that :-

pH + pOH =  14

$\sf\implies$ pH = 14 - pOH

= 14 - 3

= 11

$\sf\therefore$ pH of Ca(OH)_2 = 11

Answer 2

To Find :-

pH of $\sf {10}^{ - 3}$M solution of $\sf Ca(OH)_2$

Solution :-

We know that,

$\sf \blue{pH = -log_{10} [H^{+}]}$

$\sf \blue{pOH = -log_{10} [OH^{ - }]}$

According to question,

$\sf pOH = -log_{10} [10^{ -3}]$

$\sf pOH = -( - 3) \: log_{10} \: 10$

$\sf pOH = 3 \: log_{10} \: 10$

$\sf \fbox{pOH = 3}$

$\sf \therefore \log_a \: \fbox{a = 1}$

We also know that,

$\sf \blue{pH + pOH = 14}$

$\sf →pH = 14 - pOH$

$\sf →pH = 14 - 3$

$\sf \fbox{→pH = 11}$

$\therefore$ pH of $\sf Ca(OH)_2$ is 11.