Question

How many milliliters of 0.228 M HBrO4 are needed to titrate each of the following solutions to the equivalence point?

(a) 33.5 mL of 0.388 M KOH mL

(b) 60.1 mL of 0.205 M CsOH mL

(c) 567.0 mL of a solution that contains 2.74 g of NaOH per liter mL

Answer 1

a)

Balanced chemical equation is:

KOH + HBrO4 ---> KBrO4 + H2O

Here:

M(KOH)=0.388 M

M(HBrO4)=0.228 M

V(KOH)=33.5 mL

According to balanced reaction:

1*number of mol of KOH =1*number of mol of HBrO4

1*M(KOH)*V(KOH) =1*M(HBrO4)*V(HBrO4)

1*0.388 M *33.5 mL = 1*0.228M *V(HBrO4)

V(HBrO4) = 57.0 mL

Answer: 57.0 mL

b)

Balanced chemical equation is:

CsOH + HBrO4 ---> CsBrO4 + H2O

Here:

M(CsOH)=0.205 M

M(HBrO4)=0.228 M

V(CsOH)=60.1 mL

According to balanced reaction:

1*number of mol of CsOH =1*number of mol of HBrO4

1*M(CsOH)*V(CsOH) =1*M(HBrO4)*V(HBrO4)

1*0.205 M *60.1 mL = 1*0.228M *V(HBrO4)

V(HBrO4) = 54.0373 mL

Answer: 54.0 mL

c)

Balanced chemical equation is:

NaOH + HBrO4 ---> NaBrO4 + H2O

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass(NaOH)= 2.74 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(2.74 g)/(40 g/mol)

= 6.85*10^-2 mol

According to balanced equation

mol of HBrO4 reacted = (1/1)* moles of NaOH

= (1/1)*6.85*10^-2

= 6.85*10^-2 mol

This is number of moles of HBrO4

use:

M = number of mol / volume in L

0.228 = 6.85*10^-2/ volume in L

volume = 0.300 L

volume = 300 mL

Answer: 300 mL