Question

How many milliliters of 0.821 M HNO₃ are needed to titrate each of the following solutions to the equivalence point?




(a) 50.6 mL of 1.07 M NaOH

mL



(b) 40.0 mL of 0.657 M RbOH

mL



(c) 498.0 mL of a solution that contains 17.7 g of LiOH per liter

mL

Answer 1

a)

Balanced chemical equation is:

NaOH + HNO3 ---> NaNO3 + H2O

Here:

M(NaOH)=1.07 M

M(HNO3)=0.821 M

V(NaOH)=50.6 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of HNO3

1*M(NaOH)*V(NaOH) =1*M(HNO3)*V(HNO3)

1*1.07 M *50.6 mL = 1*0.821M *V(HNO3)

V(HNO3) = 65.9 mL

Answer: 65.9 mL

b)

Balanced chemical equation is:

RbOH + HNO3 ---> RbNO3 + H2O

Here:

M(RbOH)=0.657 M

M(HNO3)=0.821 M

V(RbOH)=40.0 mL

According to balanced reaction:

1*number of mol of RbOH =1*number of mol of HNO3

1*M(RbOH)*V(RbOH) =1*M(HNO3)*V(HNO3)

1*0.657 M *40.0 mL = 1*0.821M *V(HNO3)

V(HNO3) = 32.0 mL

Answer: 32.0 mL

c)

Balanced chemical equation is:

LiOH + HNO3 ---> LiNO3 + H2O

Molar mass of LiOH,

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

= 1*6.968 + 1*16.0 + 1*1.008

= 23.976 g/mol

mass(LiOH)= 17.7 g

use:

number of mol of LiOH,

n = mass of LiOH/molar mass of LiOH

=(17.7 g)/(23.98 g/mol)

= 0.7382 mol

According to balanced equation

mol of HNO3 reacted = (1/1)* moles of LiOH

= (1/1)*0.7382

= 0.7382 mol

This is number of moles of HNO3

use:

M = number of mol / volume in L

0.821 = 0.7382/ volume in L

volume = 0.899 L

volume = 899 mL

Answer: 899 mL