A100.0 ml sample of 0.20 m hf is titrated with 0.10 m koh. determine the ph of the solution after the addition of 200.0 ml of koh

The balanced equation for the reaction between KOH and HF is 
KOH + HF ---> KF + H₂O
stoichiometry of KOH to HF is 1:1
the number of HF moles reacted - 0.20 mol/L x 0.1000 L = 0.020 mol
the number of KOH moles reacted - 0.10 mol/L x 0.2000 L = 0.020 mol
KOH is a strong base and HF is a strong acid therefore they completely ionise into their corresponding ions
KOH --> K⁺ + OH⁻
HF ---> H⁺ + F⁻
number of H⁺ moles reacted = number of OH⁻ moles reacted
solution is completely neutralised no excess hydroxide or hydrogen ions
therefore when solution is neutral , pH = 7
pH of solution = 7

Idon't see the pic?