Exercise 7 Name Stuart Burt

Data Sample 1 Sample 2
H2C2O4 burette: Initial reading 0 mL 18.7 mL
Reading @ endpoint 18.7 mL 38.4 mL
Volume of Acid Delivered 18.7 mL 19.7 mL

NaOH Burette: Initial Reading 0 mL 18.8 mL
Reading at end-point 18.8 mL 38.3 mL
Volume of Base Delivered 18.8 mL 19.5 mL

Calculations 18.7*.1/18.8 19.7*.1/19.5
Normality of NaOH .099 N .101 N
Average Normality .1 N

NaOH Sample 1 Sample 2
Initial reading 0 mL 21.5 mL
Reading @ Endpoint 21.5 mL 43 mL
Volume Base delivered to Vinegar 21.5 mL 21.5 mL
Calculations 21.5*.1/2.5 21.5*.1/2.5
Normality of Vinegar .86 N .86 N
Calculations .86-.833/.833 .86-.833/.86
% deviation 3.24% 3.24%

Practice Problems

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
1.02/.375

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 this 4.1 eq
Solution?
150/37
C. What is the normality of this solution? 2.3 N
4.1/1.8

4.
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

5.
A. What volume of .44 N H3PO4 would be required in order to contain 2.24 equivalents of this acid?

.44X = 2.24 5.09 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

6.
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?

(60/85)/.5L .58 N

7.
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 * 36/6 540 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 N 495 mL

8.
A. How many grams of Ca(OH)2 are present in 500 mL of 1.35 N calcium hydroxide solution?

.675 * 37 25 g

B. How many grams of H4AsO4 are present in 3350 mL of .72 N solution of arsenic acid?

2.412 * 35.75 86.23 g