Calculate pH of a solution?

Question

Calculate the pH of a solution prepared by adding 115mL OF .100 M NaOH to 100mL of .100 M HNO3 solution.

I would appreciate that you explain the procedure.
Thanks in advance

Answer 1

HNO3 and NaOH react 1:1 to form water and sodium nitrate. Since you have more NaOH than HNO3, the excess hydroxide will cause the resulting solution to be basic.

You're starting with 10 mmol of HNO3 and 11.5 mmol of NaOH, so you will have 1.5 mmol of NaOH left at the end. Your total volume is 215 ml, so the [OH-] = 1.5 mmol/215 ml = 0.00698 M. pOH = 2.16, and pH = 14 - 2.16 - 11.84.

Answer 2

Before the reaction you have H+= 0.1 x 0.1 = 0.01 mol,
OH-= 0.0115 mol

NaOH + HNO3 --> NaNO3 + H2O
0.01____0.01

After the reaction, H+ = 0, OH- = 0.0015 mol
[OH-] = 0.0015/0.215 = 0.00698 (new volume is 215mL)
pOH = -log [OH-] = 2.155
pH + pOH = 14 --> pH = 11.85