Answer : The pH of the solution will be 13.63.
Solution:
Moles of hydroxide ions in barium hydroxide solution
[tex]BaOH_2\rightarrow Ba^++2(OH)^-[/tex]
Number of moles = (concentration)[tex]\times [/tex](volume in liters) ...(1)(1L=1000mL)
2.0 M barium hydroxide in solution
Number of moles of [tex]Ba(OH)_2[/tex] in 2.0 M solution[tex]=2.0 M\times 0.004 L=0.008[/tex] moles
If One mole of barium hydroxide gives two moles of hydroxide in solution then 0.008 moles will give:
Number of moles of [tex]OH^-=(2)\times (0.008 mol)=0.016[/tex] moles
Moles of [tex]H^+[/tex] ions in nitric acid
[tex]HNO_3\rightarrow NO^{-}_{3}+H^+[/tex]
In 1.00 M nitric acid solution
Moles of nitric acid in 1.0 M solution [tex]=(1.0 M)\times(0.010 L)=0.010[/tex] moles
If one mole of nitric acid will give one moles of [tex]H^+[/tex] ion.
Then 0.010 moles of nitric acid in solution will give :
Number of moles of [tex]H^+=(1)\times (0.010 mol)=0.010[/tex] moles
Since , the reaction will be neutralization reaction, equal number of moles of [tex]H^+[/tex] will neutralize equal number of moles of [tex]OH^-[/tex]
So,0.010 moles of [tex]H^+[/tex] will neutralize 0.010 moles of [tex]OH^-[/tex] in the solution
Remaining moles of [tex]OH^-=0.016- 0.01=0.006[/tex] moles
Concentration of [tex][OH^-][/tex] resulting solution
Resulting volume of the solution : 0.004 + 0.010 liters
Calculating [tex][OH^-][/tex] by using equation (1).
[tex][OH^-]=\frac{0.006}{0.004L+0.010 L}=0.4285 M[/tex]
[tex]pOH=-log[OH^-]=-log(0.4285)=0.367[/tex]
[tex]pH=14-pOH=14-0.367=13.633[/tex]
The pH of the solution will be 13.633.
