Practice problems 7b Key:
1) 3 KOH + H3PO4  3 H2O + K3PO4
!
2) a) NaOH + HNO3  H2O + NaNO3 , a 1:1 titration.
At equiv. point: # mol H+ donated =# mol H+ accepted
Thus, # mol HNO3 = # mol NaOH ; or MHNO3VHNO3 = MNaOHVNaOH
M HNO3 VHNO 3 0.500M(20.0mL)
[NaOH] = MNaOH =
=
= 0.400M
VNaOH
25.0mL
mol HNO 3,excess M HNO3 V HNO3
(.500M)(5.00mL)
b) [HNO3]excess =
=
=
=.0500M
total vol
V HNO3 + VNaOH (25.0mL + 25.0mL)
mol !
NaOH remaining
n
-n
!
c) [NaOH]
= NaOH, total NaOH, reacted =
remaining =
Vtotal
total vol
M NaOH!
VNaOH - M HNO 3 VHNO
" .500M(15.0mL)
!
!3 = .400M(25.0mL)
=
=0.0625M
(25.0mL + 15.0mL)
VHNO3 + VNaOH
0.209g 1 mol!
!
3) a) [CaSO4] =
=0.0154 M
0.100L 136.1g
b) grams undissolved
! = 0.455 – grams dissolved
= .455 – (50.0mL H2O)(.209g/100mLH2O) = 0.3505 = 0.351g
! M !V
.210M(16.5mL)
4) MHCl = NaOH NaOH =
= 0.139M
VHCl
25.0mL
5) a) H3C6H5O7+ 3 NaOH  + 3 H2O + Na3C6H5O7
!
! at equiv pt: #mol
b)
H+ donated = # mol H+ accepted
0r, # mol H3C6H5O7 x 3 = # mol NaOH
Or since n=MV: 3M1V1 = M2V2 where 1 = citric acid, 2 = NaOH
MV
0.160M(28.4mL)
So M1 = 2 2 =
=0.0473M
3V1
3(32.0mL)
c) #g H3C6H5O7 =0.0320L(0.0473mol/L)(192.1g/mol)=0.291g
! =
!d) [NaOH]
!
(5.00mL)(0.160M)
= 0.0122M
(32.0 + 28.4 + 5.00)mL