Exercise 7Name Stuart BurtDataSample 1Sample 2
H2C2O4 burette: Initial reading0 mL18.7 mL
Reading @ endpoint18.7 mL38.4 mL
Volume of Acid Delivered18.7 mL19.7 mL
NaOH Burette: Initial Reading0 mL18.8 mL
Reading at end-point18.8 mL38.3 mL
Volume of Base Delivered18.8 mL19.5 mL
Normality of NaOH.099 N.101 N
Average Normality.1 N
NaOHSample 1Sample 2
Initial reading0 mL21.5 mL
Reading @ Endpoint21.5 mL43 mL
Volume Base delivered to Vinegar21.5 mL21.5 mL
Normality of Vinegar.86 N.86 N
1. A student has a solution consisting of 8.3 g of HCl in enough water to make 700 mL of solution.
A. What is the equivalent weight of this acid?38 g/eq
B. How many equivalents of acid are present in this solution?.22 eq
8.38 g/38 g
C. What is the normality of this solution?.31 N
.22 eq/.7 L
2. A student has 32 g of H3PO4 in enough water to make 375 mL of solution.
A. What is the equivalent weight of this acid?32.6 g/eq
B. How many equivalents of this acid are present in this solution?1.02 eq
C. What is the normality of this solution?2.72 N
A student has 150 g of Ca(OH)2 in enough water to make 1800 ml of solution.
A. What is the equivalent weight of this base?37 g/eq
B. How many equivalents of this base are present in this4.1 eq
C. What is the normality of this solution?2.3 N
A. If 70 mL of a NaOH solution required 44 mL of .48 N H2SO4 to titrate it, what
70 mL * X = 44 mL * .48 N.3 N
B. What would be the normality of a sodium hydroxide solution if 39 mL of it required 2.25 g of H2SO4 to titrate it?
39 mL * X = 44mL * (2.25/49)/44)1.2 N
A. What volume of .44 N H3PO4 would be required in order to contain 2.24 equivalents of this acid?
.44X = 2.245.09 L
B. What volume of 1.5 N H2SO4 would be needed in order to contain 13 g of this acid?
1.5X= (13/49).18 L
A. What is the concentration of the resulting solution if 400 mL of .68 N HCl was diluted to 750 mL?
400 * .68 N/750 N.36 N
B. What is the concentration of the resulting solution when 60 g of Ba(OH)2 in 500 mL of solution is diluted to 1200 mL?
A. To what volume would you have to dilute 90 mL of 36 N H2SO4 so that the resulting solution had a concentration of 6 N?
90 mL * 36/6540 mL
B. To what volume would you have to dilute 33 mL of 6 M HCl in order for the resulting solution to have a concentration of .4 N?
33 mL *6 N /4 N495 mL
A. How many grams of Ca(OH)2 are present in 500 mL of 1.35 N calcium hydroxide solution?
.675 * 3725 g
B. How many grams of H4AsO4 are present in 3350 mL of .72 N solution of arsenic acid?
2.412 * 35.7586.23 g