Question

Calculate pH at the beginning and endpoint of the reaction with 0.2600 g KHP in 25.00 mL

of water with 0.100 M NaOH.

Answer 1

Moles of KHP = Mass/MW = 0.26/204.22 = 0.00127

Initial conc of acid, C = Moles/Volume = 0.00127/0.025 = 0.0508 M

Ka for HP = 10^-5.4.

So,

Initial [H⁺] = (Ka*C)^0.5 = (10^-5.4*0.0508)^0.5 = 0.00045

So,

Initial pH = -log(0.00045) = 3.346

Volume of base required to reach end point = Moles of acid/Molarity of base = 0.00127/0.1 = 0.0127 L = 12.7 mL

So,

Final volume = 25+12.7 = 37.7 mL = 0.0377 L

So, conc of salt formed at end point, C' = 0.00127/0.0377 = 0.0337 M

Kb = 10^-14/Ka = 10^-14/10^-5.4 = 10^-8.6

Final [OH-] = (Kb*C')^0.5 = (10^-8.6*0.0337)^0.5 = 0.0000092

So,

Final pOH = -log(0.0000092) = 5.036

Final pH = 14-pOH = 14-5.036 = 8.964

Hope this helps !