Question

What is the pH of a solution in which 25.0 mL
of 0.1 M NaOH is added to 25 mL of 0.08 M
HCl and final solution is diluted to 500 mL ?
(A) 3
(B) 11
(C) 12
(D) 13​

Answer 1

Answer =11

NaOH+HCl=>NaCl+H2O

25ml of 0.1M NaOH

Here,no.of moles of NaOH=0.0025(from Mv=n)

Similarly no.of moles of HCl =0.002

Clearly from the chemical eqn,-1 mole of NaOH reacts with 1 mole of HCl

Therefore,0.002 moles of HCl reacts with 0.002 moles of NaOH,which means 0.0005moles of NaOH is remaining

Molarity of remaining NaOH=(0.0005÷500)×1000

Molarity =0.001

Poh of 0.001M solution=-log(0.001)

Poh=3

Therefore,Ph=14-3=11

XY=11=> X+Y=2