Question

9. 100 mL of 0.01 M solution of NaOH is diluted to
1 litre. The pH of resultant solution will be
​

Answer 1

Answer: 11

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock $NaOH$solution = 0.01 M

$V_1$ = volume of stock $NaOH$ solution = 100 ml

$M_2$ = molarity of diluted $NaOH$ solution = ?

$V_2$ = volume of diluted $NaOH$ solution = 1 L =1000 ml

$0.01M\times 100=M_2\times 1000$

$M_2=0.001$

The concentration of $OH^-$ ions is 0.001 M

pH or pOH is the measure of acidity or alkalinity of a solution.

pOH is calculated by taking negative logarithm of hydroxide ion concentration.

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

Putting in the values:

$pOH=-\log[0.001]$

$pOH=3$

$pH+pOH=14$

$pH=14-3=11$

Thus pH is 11.

Answer 2

Answer:

hope this helps you

pl mark as brainliest!!!!