Question

A student is titrating 50.00 mL of a 0.100 M weak acid (Ka = 1.8x10-5) with 0.100 M KOH. What is the pH after 30.00 mL of the 0.100 M KOH solution has been added?

Answer 1

Let the weak acid be HA, volume - 50 ml = 0.650 L . Moles of weak acid = volume in LX concentration = 0.050 % 0.1 Moles of KOH consumed =0:130 X 0.1 The chemical ran is , HA +KOH KA + H₂O Acid Base 1. So, 0.030x0.1 moles of kon reacts with 0.0300 0-1 moles of HA to pm диа 0- 030ҳ0-1 mglau of KA. - Remaining moles of HA = (0.050*0.1 -0.03000.1) = 0.020 XO1 Now, Applying Henderson Equation we can find pH as follows moles of salt pH = -log Ka + log me if acid = -log (18x175) + log 0.030X0.1 = 4.74 + log 2/2 = 4.74 + 0.239 = 4.979 So, pH after one of 0.1 M KOH added = 4.979 0.020 X 0.1

For any confusion feel free to ask.

Do rate if you got it.

Thanks,