Question

A student is asked to standardize a solution of barium hydroxide. He weighs out 1.06 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 23.0 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution? =M

This barium hydroxide solution is then used to titrate an unknown solution of hydrochloric acid. B. If 20.7 mL of the barium hydroxide solution is required to neutralize 12.8 mL of hydrochloric acid, what is the molarity of the hydrochloric acid solution?=M

Answer 1

monoprotic means it donates one proton or H⁺ ion for neutralization of 0.5 MOLE of Ba(OH)₂

1. Write and balance the equation. If we call potassium hydrogen phthalate just KHP, then
Ba(OH)2 + 2KHP ==> 2H2O + BaP + K2P

2. Convert grams to moles. given mass of KHP =1.06g, molar mass of KHP= 204

moles of KHP = 1.06/molar mass

KHC8H4O4 = 1.06/204 = approximately 0.0052

3. Using the coefficients in the balanced equation, convert moles KHP to moles Ba(OH)2
0.0052moles KHP x (0.5)= 0.0052 x (1/2) = 0.0026 moles Ba(OH)₂

4. M of Ba(OH)₂ = moles Ba(OH)₂/ volume of  Ba(OH)₂ in L

volume given =23 ml= 0.023 L
M Ba(OH)2 = 0.0026/0.023= approximately 0.113 M

so the molarity of the barium hydroxide solution = 0.113 M

B).

using above M of the barium hydroxide solution = 0.113 M , and volume given= 20.7 ml= 0.0207 L

to neutralize 12.8 ml of HCl. so again compare the gram equivalent of both using balance chemical reaction as below:

Ba(OH)2 + 2HCL ==> 2H2O + Bacl_2(aq)

so we need 0.5 mol of Ba(OH)2 to neutralize the 1 mole of HCL.

now find N1= moles of Ba(OH)2 = M1*V1= 0.113*(0.0207)

N1=0.00234 moles

so 2.34 moles will neutralize the 2*(0.00234) (double)mole of HCL

so N2= 2*(0.00234) = 0.00468 MOLES of HCL

now use N2= M2*V2

given V2=12.8 ml =0.0128 L , N2= 0.00468 So

M2= 0.00468/(0.0128)

M2=0.365 M

so 0.365 M HCl