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?

13.63
0.37
0.85
13.15

Answer 1

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:

`2HNO_3(aq)+Ba(OH)_2(aq)\rightarrow Ba(NO_3)_2(aq)+2H_2O(l)`

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

moles of barium hydroxide =
`4.0mL((1L)/(1000mL))((2.0mol)/(1L))`

= 0.008 mol

moles of nitric acid =
`10.0mL((1L)/(1000mL))((1.00mol)/(1L))`

= 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:

`0.01molHNO_3((1molBa(OH)_2)/(2molHNO_3))`

=
`0.005molBa(OH)_2`

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 =
`(0.003mol)/(0.014L)`

= 0.214M

`Ba(OH)_2(aq)\rightarrow Ba^2^+(aq)+2OH^-(aq)`

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

So,
`[OH^-]=2*0.214M` = 0.428M

`pOH=-log[OH^-]`

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.