Question

4. Knowing that one mole of KHP, C3H5O4K, reacts with one mole of NaOH, what mass of KHP is required to neutralize 30.0 mL of the 0.10 M NaOH solution? 5. If 24.5 mL of the 0.10 M NaOH solution is required to reach the endpoint in a titration with an unknown monoprotic acid, how many moles of the acid were present?

Answer 1

(4) KHP + NaOH KNAP + H₂O 0.1M NaOH Millimeles of NaOH in 30mL of = Concentration & Volume = 0.1m x 30 mL = 3 mmeles 1 mmel = 10 3 mel. 3mmeles - 3 x 103 moles. Meles of NaOH = 3x10-3 meles. 1 more of Naoh reacts with 1 mel. KHP 3x10 3 meles Naon reacts with ind KHP x 3x10 mel Imol NaOH NOOH - 3x103 ma kHP. Molar mass of KHP (C8H5O4K) = 204-22 glmel Mass of KHP = Meles & Molar mass. 3x103 ml x 204. 22g mel Mass of KHP = 0.613 g 9 of is neutralize 0.613 30mL KHP O IM required to NaOH solution

(5) HA + NaOH - NAA + H₂O. Mono protic Acid At the end point of titration & Moles of acid equal to the moles of base. Moles of monoprotic acid - Moles of Naoh Millimees of NaOH in 24.5 me of 0.1M NaOH = O.IM x 24. 5mL = 2.45 mmel. Moles of NaOH = 103 mal x 2.45 mmol. Immel - 2:45. X10-3 mells of NoBH | Moles of Acid = 2.45 x 10-3 moles. OR 2.45 x 103 meles of the acid were present.

(5) HA + NaOH - NAA + H₂O. Mono protic Acid At the end point of titration & Moles of acid equal to the moles of base. Moles of monoprotic acid - Moles of Naoh Millimees of NaOH in 24.5 me of 0.1M NaOH = O.IM x 24. 5mL = 2.45 mmel. Moles of NaOH = 103 mal x 2.45 mmol. Immel - 2:45. X10-3 mells of NoBH | Moles of Acid = 2.45 x 10-3 moles. OR 2.45 x 103 meles of the acid were present.