Question

$\bold{chemistry \: question}$
$\sf{calculate \: the \: pH \: of \: a \: {10}^{ - 3} M \: solution \: of \: Ca(OH) \tiny{2} \small{ \: if \: it \: undergoes \: complete \: ionisation}}\ \textless \ br /\ \textgreater \ ​$
$\star\sf\red{\: standard - 12 \: th}$
$\star \sf \red{ \: chapter - ionic \: equilibria}$
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Answer 1

Given to calculate pH of a compound :-

${Ca(OH)_2}$ where equal to ${10^{-3}}$

SOLUTION:-

As ${Ca(OH)_2}$ is a base then ${OH^-}$ ions present So,

$OH {}^{ - } = 10 {}^{ - 3}$

then ,

$pOH = \: - log(oh {}^{ - } )$

$pOH = - log(10 {}^{ - 3} )$

It is in form of ,

$log(a {}^{m} ) = m log(a)$

So,

$pOH = \: - ( -3 ) log(10)$

$pOH = 3 log(10)$

Here base anything not there that means Its base is 10

So,

$log_{a}(a) = 1$

$pOH = 3(1)$

$pOH {}^{ } \: = 3$

As we know that ,

$pH + pOH = 14$

$pH + 3 = 14$

$pH = 14- 3$

$pH = 11$

So, pH of ${Ca(OH)_2}$ is 11

Answer 2

Answer:- pH = 12.3

Answer:- M/100 calcium hydroxide means 0.01M solution. Calcium hydroxide is a strong base and breaks to give hydroxide ions in aqueous solution as:

Ca(OH)_2\ Ca^+^2+2OH^-

From this equation, 1 mol of calcium hydroxide gives 2 moles of hydroxide ion. So, the concentration of hydroxide ions would also be two times to the given concentration of calcium hydroxide.

• pOH is calculated using the formula:

• pOH=-log[OH^-]pOH=−log[OH− ]

pOH =- log(0.02)

pOH = 1.7

Now, we could calculate the pH using the formula:

• pH = 14 - pOH

• pH = 14 - 1.7
• pH = 12.3

• So, the pH of M/100 calcium hydroxide solution is 12.3.