What is the pH of the resulting solution when we add 0.040 mol of

Question

𝐾𝑂𝐻 to 1.0 L of 0.10 M carbonic acid 𝐻2𝐶𝑂3 𝐾𝑎 = 6.2 × 10−7?

Answer 1

I'm assuming this is meant to be a buffer problem.
.............H2CO3 + KOH ==> KHCO3 + H2O
I..............0.1 mol....0...................0............0
add........................0.04.........................
C............-0.04.......-0.04..........+0.04
E..............0.06..............0...............0.04
Plug the E line into the Henderson-Hasselbalch equation
Post your work if you get stuck.