Question

If 10.0 mL of 0.20 M NaOH is added to 50.0 mL of 0.10 M HCl, what will be the pH of the resulting solution?

Answer 1

NaOH= 10.0ml of 0.20M

number of moles of NaOH= 0.20M x 0.010L= 0.002 moles

HCl= 50.0 mL of 0.10M

number of mole of HCl = 0.1M x0.050L= 0.005 moles

number of moles of HCl is more than that of NaOH

So , the resultant solution has acidic in nature.

Remaining number of moles of Acid = 0.005 - 0.002 = 0.003 moles

number of moles of H+ ion = 0.003 moles

Total volume of the solution = 10.0+50.0= 60.0mL= 0.060L

Concentration of the resultant solution = number of moles/volume in L = 0.003/0.060= 0.05M

[H+] = 0.05M

-log[H+]= -log(0.05)

PH= 1.30