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 : 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)
`*`(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* 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)* (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)*(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)* (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^-]=(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.