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? 13.63 0.37 0.85 13.15

- 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:

let's calculate the moles of each from given molarities and mL.

moles of barium hydroxide =

= 0.008 mol

moles of nitric acid =

= 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:

=

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 =

= 0.214M

Barium hydroxide as two OH in it. So, the concentration of hydroxide ions will be twice of barium hydroxide concentration.

So,  = 0.428M

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.