Solution to Exercises for Study Sheet 16
NOTE: A balanced chemical equation is needed when doing the stoichiometric calculations in this
study sheet involving neutralization reactions (i.e., the reaction of an acid with a base).
1. 3 Ba(OH)2 (aq)
+ 2 H3PO4 (aq)
→
Ba2(PO4)2 (s)
For the sake of convenience - let, Ba(OH)2 = MOH
? 𝑚𝐿 𝑀𝑂𝐻 = 22.4 𝑚𝐿 𝑀𝑂𝐻 𝑠𝑜𝑙𝑛
2. 2 NaOH (aq)
+ 1 H2C2O4 (s)
→
+
6 H2O (𝓁)
and H3PO4 = HA
!.!"! !"# !"#
! !"# !"
!""" !" !" !"#$
!""" !" !"# !"#$
! !"# !"#
!.!"# !"# !"
0.121 M Ba(OH)2
coefficients
of bal. eq.
Na2C2O4 (aq)
+
= 58.3 𝑚𝐿
0.120 M H3PO4
2 H2O (𝓁)
For the sake of convenience - let, H2C2O4 = HA (MM = 90.0)
? 𝑚𝑜𝐿 𝑁𝑎𝑂𝐻 = 0.214 𝑔 𝐻𝐴
? MNaOH =
3. NaOH (aq)
NaOH (aq)
!.!" ! !"!! !"# !"#$
!.!"#$ ! !"#$ !"#$
! !"# !!
! !"# !"#$
!".! ! !"
! !"# !"
MMHA
coefficients
of bal. eq.
= 0.173 𝑀 + HCl (aq) → NaCl (aq) + H2O (𝓁) (1st titration)
+ HA (aq) → NaA (aq) + H2O (𝓁) (2nd titration) where, HA ⇒ monoprotic acid
? 𝑚𝑜𝐿 𝑁𝑎𝑂𝐻 = 25.0 𝑚𝐿 𝐻𝐶𝑙 𝑠𝑜𝑙𝑛
? MNaOH =
= 4.76 𝑥 10!! 𝑚𝑜𝑙
!.!" ! !"!! !"# !"#$
!.!!"# ! !"#$ !"#$
!.!"# !"# !"#
! !"# !"#$
!""" !" !"# !"#$
! !"# !"#
0.192 M HCl
coefficients
of bal. eq.
= 0.113 𝑀 ? 𝑚𝑜𝐿 𝐻𝐴 = 32.4 𝑚𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑛
!.!!" !"# !"#
! !"# !"#$
!""" !" !"# !"#$
! !"# !"#
? MMHA =
4.
!.!! ! !"!! !"# !"
= 3.66 𝑥 10!! 𝑚𝑜𝑙
coefficients
of bal. eq.
0.113 M HCl
!.!"# ! !"
= 4.80 𝑥 10!! 𝑚𝑜𝑙
= 112 𝑔/𝑚𝑜𝑙
Sr(OH)2 + 2 KHP → SrP + K2P + 2 H2O
Sr(OH)2 + 2 HC2H3O2 → Sr(C2H3O2)2 + 2 H2O (𝓁)
(1st titration)
(2nd titration)
For the sake of convenience - let, Sr(OH)2 = MOH and HC2H3O2 = HA
? 𝑚𝑜𝐿 𝑀𝑂𝐻 = 0.613 𝑔 𝐾𝐻𝑃
? MMOH =
!.!" ! !"!! !"# !"#
!.!"#$ ! !"#$ !"#$
! !"# !"#
! !"# !"#
!"#.! ! !"#
! !"# !"#
MMKHP
coefficients
of bal. eq.
= 1.50 𝑥 10!! 𝑚𝑜𝑙
= 0.0858 𝑀
? 𝑔 𝐻𝐴 = 31.4 𝑚𝐿 𝑀𝑂𝐻 𝑠𝑜𝑙𝑛
!.!"#" !"# !"#
! !"# !"
!".! ! !"
!""" !" !"# !"#$
! !"# !"#
! !"# !"
0.0858 M MOH
coefficients
of bal. eq.
MMHA
= 0.323 𝑔
% HA =
!.!"! ! !"
!.!" ! !"#$%& !"#$%& • 100 = 4.59 % 4. KHP + NaOH → KNaP + H2O
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
For the sake of convenience - let, H2C2O4 = HA
? 𝑚𝐿 𝑁𝑎𝑂𝐻 = 0.124 𝑔 𝐾𝐻𝑃
required to
neutralize KHP
? 𝑚𝐿 𝑁𝑎𝑂𝐻 = 0.140 𝑔 𝐻𝐴
required to
neutralize HA
! !"# !"#
! !"# !"#$
!""" !" !"#$ !"#$
!"#.! ! !"
! !"# !"#
!.!"# !"# !"#$
MMKHP
coefficients
of bal. eq.
MNaOH
! !"# !"
! !"# !"#$
!""" !" !"#$ !"#$
!".! ! !"
! !"# !"#
!.!"# !"# !"#$
MMKHP
coefficients
of bal. eq.
= 5.8 𝑚𝐿
= 30.0 𝑚𝐿
MNaOH
? mL NaOH required to neutralize both KHP and HA = 5.8 mL + 30. 0 mL = 35.8 mL