Question

QUESTIONS (Show your work) 1. 1, In a titration experiment, 0.250 M KOH is added from a buret to 27.4 mL of a 0.154 M HSO, solution and reacts according to the equation: H;SO. (aq)- 2 KOH (aq) -- K.SO. (aq)+ 2 HO (aq) How many milliliters of potassium hydroxide solution have been added at the equivalence point (end point)? 2. In a titration experiment,, 0.375 M NaOH is added from a buret to 31.2 mL of a 0.482 M H,PO4 solution and reacts according to the equation: 3 NAOH (aq) + HPO. (aq)-------NasPOs (aq) + 3 H2O (aq) How many milliliters of sodium hydroxide solution have been added at the equivalence point (end point)?

Answer 1

1)

Balanced chemical equation is:

H2SO4 + 2 KOH ---> K2SO4 + 2 H2O

Here:

M(H2SO4)=0.154 M

M(KOH)=0.25 M

V(H2SO4)=27.4 mL

According to balanced reaction:

2*number of mol of H2SO4 =1*number of mol of KOH

2*M(H2SO4)*V(H2SO4) =1*M(KOH)*V(KOH)

2*0.154 M *27.4 mL = 1*0.25M *V(KOH)

V(KOH) = 33.8 mL

Answer: 33.8 mL

2)

Balanced chemical equation is:

H3PO4 + 3 NaOH ---> Na3PO4 + 3 H2O

Here:

M(H3PO4)=0.482 M

M(NaOH)=0.375 M

V(H3PO4)=31.2 mL

According to balanced reaction:

3*number of mol of H3PO4 =1*number of mol of NaOH

3*M(H3PO4)*V(H3PO4) =1*M(NaOH)*V(NaOH)

3*0.482 M *31.2 mL = 1*0.375M *V(NaOH)

V(NaOH) = 120. mL

Answer: 120. mL