Chemistry 11 Hand In Assignment # 11 – Molarity, Excess and % Yield Problems

Hand In Assignment # 11 – Molarity, Excess and % Yield Problems Page 1 of 2

Name ___________________________ Due Date ________________________ Mark ________/17

Chemistry 11

Hand In Assignment # 11 – Titration Check the number of SF according to the rule. Show all work to get full mark.

1. Given the following neutralization reaction, answer the questions below it.

Ca(OH)2(aq) + 2 HCl(aq) ! CaCl2(aq) + 2 H2O(l)

If 16.50 mL of 0.200 M Ca(OH)2 is required to reach equivalence point with 25.0 mL of a solution of HCl, find the [HCl]. (4 marks)

Answer _______________________

What volume of 0.200 M Ca(OH)2 in mL would be required to titrate 250 mL of 0.0365 M of HCl? (4 marks)

Answer _______________________

8

HCI CALOHK

Cone . ( M ) ? 0,200

VOLUME ( ml ) 25,0 16,50

✓ ✓ ✓[ Hu ]= 16.50mL

-101200M¥CALOH)z× KHU I = 0.264 MHCI

x -

1mA CALOHK 25.0mL

0.264N 3sF 14

CALOHK HCI

Volume ( ml ) ? 250

Cone . ( M ) 01200 0,0365

✓ ✓ ✓HU 1 mol CALOHK

VCACOH )z= 250mL -10.0365M¥ - × -2=22 .8125mL

QMOIHCI01200 mol

23mL 2SF 14

Chemistry 11 Hand In Assignment # 11 – Molarity, Excess and % Yield Problems

Hand In Assignment # 11 – Molarity, Excess and % Yield Problems Page 2 of 2

2. Given the following neutralization equation, answer the questions below it.

__ Ca(OH)2(aq) + __ H2SO4(aq) ! _____________ +_____________

a. Predict the product and balance the equation. (1 mark) b. To titrate 0.0250 M of 10.0 mL calcium hydroxide, 13.5 mL of sulfuric acid was

required to reach equivalance point. Calculate the concentration of the acid. (4 marks)

Answer _______________________

c. Stephen titrated 10. mL of 2.3 x 10-2 M sulfuric acid with calcium hydroxide, and calculated the concentration of calcium hydroxide as 5.0 x 10-3 M. However, he lost his data table which had the volume of calcium hydroxide written. Calculate the volume of calcium hydroxide needed to reach equivalence point in mL. (4 marks)

Answer _______________________

9

2h20

Cas04✓CALOHK HZSO

.4Cone CM ) 0,0250 ?

Volume ( ML ) 10,0 13.5

✓ 1mm

#a

✓[ HzsO4 ] = 10.0mL XOIOZBOMOICALOHK

- ×-1

I ×1 mol Catlett )z 13.5mL

=

0.01852N 0.0185N 3SF 15

CALOH )2 H2S04

Cone .C M ) 5,0×10-3 2.3×10-2

Volume ( ML ) ? 10 .

✓ ✓VCALOH )z= 10 .mn/2i3XlJ2mo1HzsOqlmo1Cal0H)z

✓±T

× 1mo#zsO¢×5,0×10 -3mg

= 46mL 46mL 25+14