Question

Determine the pH of a solution when 25.7 mL of 0.13 M HNO3 is mixed with 20.5 mL of:

 

A) 0.100 M NaOH

B) Distilled water

C) .060M HCl
D).300M KOH

 

Answer 1

Moles of HNO3 = volume x concentration of HNO3

= 25.7/1000 x 0.13 = 0.003341 mol


Total volume = 25.7 + 20.5 = 46.2 mL = 0.0462 L


(A) Moles of NaOH = volume x concentration of NaOH

= 20.5/1000 x 0.100 = 0.00205 mol


NaOH + HNO3 => NaNO3 + H2O

Excess moles of HNO3 = 0.003341 - 0.00205 = 0.001291 mol


[H+] = [HNO3] = excess moles of HNO3/total volume

= 0.001291/0.0462 = 0.02794 M


pH = -log[H+] = -log(0.02794) = 1.55


(B) [H] = [HNO3] = moles of HNO3/total volume

= 0.003341/0.0462 = 0.07232 M


pH = -log[H+] = -log(0.07232) = 1.14


(C) Moles of HCl = volume x concentration of HCl

= 20.5/1000 x 0.060 = 0.00123 mol


[H+] = (moles of HNO3 + moles of HCl)/total volume

= (0.003341 + 0.00123)/0.0462 = 0.09894 M


pH = -log[H+] = -log(0.09894) = 1.00


(D) Moles of KOH = volume x concentration of KOH

= 20.5/1000 x 0.300 = 0.00615 mol


KOH + HNO3 => KNO3 + H2O

Excess moles of KOH = 0.00615 - 0.003341 = 0.002809 mol


[OH-] = [KOH] = excess moles of KOH/total volume

= 0.002809/0.0462 = 0.06080 M


pOH = -log[OH-] = -log(0.06080) = 1.22

pH = 14 - pOH = 14 - 1.22 = 12.78