Hey there!:
Concentration of NaOH = 0.200 M
Concentration of HNO₃= 0.200 M
Total volume = 50.0 mL + 60.0 mL = 110 mL=> 0.11 L
The neutralization reaction between NaOH and HNO3 :
OH⁻ + H⁺ ----------> H₂O
So :
n ( H⁺ ) = 60 mL * 0.200 M / 1000 mL => 0.012 moles of H⁺
n ( OH⁻ ) = 50 mL 0.200 M / 1000 mL => 0.01 moles of OH⁻
Hence OH⁻ is limiting reagent .
Remaining moles of H⁺ = 0.012 - 0.01 => 0.002 moles
Concentration of H⁺ = 0.002 M / 0.11 L
Concentration of H⁺ = 0.01818 moles/L
Therefore:
pH = - log [ H⁺ ]
pH = - log [ 0.01818 ]
pH = 1.74
Hope that helps!
