How to find the pH of 32 mL of 0.1 M of NaOH added to 100mL of 0.05 of NaHCO3
Steps please!!!
OH- + HCO3 - ---> CO2 - +H2O
2.4x 10^-2 3.78 x 10^-2 0
-x -x +x
4.68 x 10^-11 = [2.4x 10^-2-x][3.78 x 10^-2-x]/
x = 2.06 x 10^-7
There are no H3O+ or OH -, how do I find the pH???