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.