Answer:- pH sis 13.63
Solution:- It is a strong base vs strong acid titration. The equation for the reaction takes place between given acid and base is:
[tex]2HNO_3(aq)+Ba(OH)_2(aq)\rightarrow Ba(NO_3)_2(aq)+2H_2O(l)[/tex]
let's calculate the moles of each from given molarities and mL.
moles of barium hydroxide = [tex]4.0mL(\frac{1L}{1000mL})(\frac{2.0mol}{1L})[/tex]
= 0.008 mol
moles of nitric acid = [tex]10.0mL(\frac{1L}{1000mL})(\frac{1.00mol}{1L})[/tex]
= 0.01 mol
From balanced equation base and acid react in 1:2 mol ratio. So, let's calculate the moles of base used to react with the acid:
[tex]0.01molHNO_3(\frac{1molBa(OH)_2}{2molHNO_3})[/tex]
= [tex]0.005molBa(OH)_2[/tex]
excess moles of barium hydroxide = 0.008 - 0.005 = 0.003 mol
Total volume of the solution = 0.004L + 0.010 mL = 0.014 L
Concentration of excess barium hydroxide = [tex]\frac{0.003mol}{0.014L}[/tex]
= 0.214M
[tex]Ba(OH)_2(aq)\rightarrow Ba^2^+(aq)+2OH^-(aq)[/tex]
Barium hydroxide as two OH in it. So, the concentration of hydroxide ions will be twice of barium hydroxide concentration.
So, [tex][OH^-]=2*0.214M[/tex] = 0.428M
[tex]pOH=-log[OH^-][/tex]
pOH = log(0.428)
pOH = 0.37
pH = 14 - pOH
pH = 14 - 0.37
pH = 13.63
First choice is correct, the pH of the solution is 13.63.
