Question

6 g koh is dissolved in 100l of solution. Calculate ph at 353 k if kw=10-12

Answer 1

Answer : The pH is, 9.03

Solution :

First we have to calculate the molarity of KOH solution.

$Molarity=\frac{\text{Moles of KOH}}{\text{Molar mass of KOH}\times \text{volume of solution}}=\frac{6g}{56g/mole\times 100L}=1.07\times 10^{-3}mole/L$

As KOH dissolves in water it dissociates into ions.

$KOH\rightarrow K^++OH^-$

From this we conclude that,

The concentration of $OH^-$ ion = concentration of KOH = $1.07\times 10^{-3}mole/L$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.07\times 10^{-3})=2.97$

Now we have to calculate the pH.

$pH+pOH=pKw$

$pH+2.97=12$

$pH=9.03$

Therefore, the pH is, 9.03