Question

A solution is made by adding $5.6$ grams of $KOH$ (molar mass $56$ grams) to enough water to make 1.0 liter of solution. What is the approximate pH of the resulting solution?

• A. $1$
• B. $3$
• C. $7$
• D. $9$
• E. $13$

Answer 1

Answer: (E)

$13$

Given $5.6$ gm of $KOH$ $\therefore n=\left(\frac{5.6}{56}\right)=0.1$ mole $KOH$

Molarity $=\left(\frac{0.1}{1.0}\right)=0.1M$ $KOH$

$KOH\to K^{+}+{OH}^{-}(\because strongelectrolyte)$

$\therefore \left[{OH}^{-}\right]=0.1M$

$\therefore pOH=-log\left({10}^{-1}\right)=1;(pH+pOH=14)$

$(\therefore pH=13)$