What weight of potassium acid phtalate KHP is required for n

What weight of potassium acid phtalate (KHP) is required for neutralization of 45.0 mL of 0.095 M NaOH?

Answer

Mass of KHP required = 0.8730g

Explanation

Reaction between NaOH and KHP is

KHP + NaOH - - - - - -> NaKP + H2O

stoichiometrically, 1mole of NaOH react with 1mole of KHP

No of moles of NaOH =( 0.095mol/1000ml)×45ml =0.004275

0.004275moles of NaOH need 0.004275moles of KHP

Molar mass of KHP = 204.22g/mol

Mass of KHP required = 0.004275mol × 204.22g/mol =0.8730g