5 questions pertaining to normality, and gram equivalent mass?

Question

Any help is appreciated, and needed.

1. What volume of 0.800M KOH is needed to neutralize 15.0mL of 0.65M H2SO4?

2. What is the molarity of sodium hydroxide if 38mL of the solution is titrated to the end point with 14mL of 0.75M sulfuric acid?

3. It requires 24.6 mL of Ca(OH)2 solution to neutralize 14.2 mL of 0.0140M HCH3COO. What is the concentration of the calcium hydroxide solution?

4. By titration, it is found that 12.4 mL of H2SO4 is required to neutralize 19.8 mL of 0.100M Ca(OH)2. What is the molarity of the H2SO4?

5. What is the volume of 0.120M Ba(OH)2 is required to neutralize 12.2 mL of 0.25M HCl?

Like I said, any help is appreciated, and needed. I have no hope of passing this class unless I get an understanding of this subject.

Answer 1

1.
Moles H2SO4 = 15.0 x 0.65 /1000 = 0.00975
Moles H+ = 2 x 0.00975 = 0.0195 = moles OH-
M = moles / Volume
Volume = moles / M = 0.0195 / 0.800 = 0.0244 L

2.
Moles H2SO4 = 14 x 0.75 /1000 = 0.0105
moles H+ = 2 x 0.0105 = 0.0210 = moles OH-
38 mL = 0.038 L
M = 0.0210 / 0.038 = 0.553 M

3.
Moles CH3COOH = 14.2 x 0.0140 /1000 = 0.000199
Moles Ca(OH)2 = 0.000199 / 2 = 0.0000994
24.6 mL = 0.0246 L
M = 0.0000994 / 0.0246 = 0.00404 M

4.
Moles H2SO4 = 19.8 x 0.100 / 1000 = 0.00198 =
moles Ca(OH)2 since the ratio between Ca(OH)2 and H2SO4 is 1 : 1
M = 0.00198 / 0.0124 L = 0.160 M

5.
Moles HCl = 12.2 x 0.25 /1000 = 0.00305
Moles Ba(OH)2 = 0.00305 / 2= 0.00153
V = 0.00153 / 0.120 = 0.0127 L

Answer 2

1) H2SO4+KOH---> Ka2SO4+2H2O
so you see KOH has a valence 1 and H2SO4 valence 2
1 L of 0.65M H2SO4 corresponds to 0.650M and 15 mL corresponds to 9.75mM, as H2SO4 has a valence 2 this corresponds to 19.5mEq. A liter of 0.800M KOH corresponds to
800mM= 800Eq since KOH has valence 1, 1mL corresponds to 0.8mEq. so to equal 19.5mEQ you need 19.5/0.8=24.37mL

2)As previously , you need twice as much NaOH that H2SO4
so 14mL of H2SO4 corresponds to 28mL NaOH.
you must write than c*38=28*0.75 and c= 0.552M NaOh

3) Ca(OH)2 corresponds to 2 HCH3COOso 24.6*2=49.2
and 49.2*c= 14.2*0.014 =0.004Eq Ca(OH)2=0.002mole Ca (OH)2 by liter

Answer 3

1) for KOH 0.800 M = 0.800 N

For H2SO4 0.65 M = 1.30 N

Na*Va = Nb*Vb

Vb = Na*Va / Nb = 1.3 * 15 / 0.800 = 24.38 ml ans

2) 0.75 M = 1.50 N


N of NaOH = 14 * 1.50 / 38 = 0.60 N

for NaOH ; M = N = 0.60 N ans

3) for acetic acid 0.014 M = 0.014 N

N of Ca(OH)2 = 14.2 * 0.014 / 24.6 = 0.008 N or 0.004 M

4 ) for Ca(OH)2 0.100 M = 0.200 N

N of H2SO4 = 0.200* 19.8 / 12.4 = 0.319 N

M of H2SO4 = 0.319 /2 = 0.051 M

5) for HCl ; 0.25 M = 0.25 N

for Ba(OH)2 = 0.120 M = 0.12*2 = 0.24 N

Vb = 12.2 * 0.25 / 0.24 = 12.71 ml of Ba(OH)2