Question

What is pH when 4.0 mL of 2.0 M barium hydroxide is added to 10.0 mL of 1.00 M nitric acid?

Answer 1

Answer : The pH of the solution will be 13.63.

Solution:

Moles of hydroxide ions in barium hydroxide solution

$BaOH_2\rightarrow Ba^++2(OH)^-$

Number of moles = (concentration)$\times$(volume in liters) ...(1)(1L=1000mL)

2.0 M barium hydroxide in solution

Number of moles of $Ba(OH)_2$ in 2.0 M solution$=2.0 M\times 0.004 L=0.008$ moles

If One mole of barium hydroxide gives two moles of hydroxide in solution then 0.008 moles will give:

Number of moles of  $OH^-=(2)\times (0.008 mol)=0.016$ moles

Moles of $H^+$ ions in nitric acid

$HNO_3\rightarrow NO^{-}_{3}+H^+$

In 1.00 M nitric acid solution

Moles of nitric acid in 1.0 M solution $=(1.0 M)\times(0.010 L)=0.010$ moles

If one mole of nitric acid will give one moles of $H^+$ ion.

Then 0.010 moles of nitric acid in solution will give :

Number of moles of  $H^+=(1)\times (0.010 mol)=0.010$ moles

Since , the reaction will be neutralization reaction, equal number of moles of $H^+$ will neutralize equal number of moles of $OH^-$

So,0.010 moles of $H^+$ will neutralize 0.010 moles of $OH^-$ in the solution

Remaining moles of $OH^-=0.016- 0.01=0.006$ moles

Concentration of $[OH^-]$ resulting solution

Resulting volume of the solution : 0.004 + 0.010 liters

Calculating $[OH^-]$ by using equation (1).

$[OH^-]=\frac{0.006}{0.004L+0.010 L}=0.4285 M$

$pOH=-log[OH^-]=-log(0.4285)=0.367$

$pH=14-pOH=14-0.367=13.633$

The pH of the solution will be 13.633.