CHE 102LLR Study Guide - Midterm Guide: Molar Mass, Titration, Ion

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coffeeserval946

31 May 2018

School

University at Buffalo

Department

Chemistry

Course

CHE 102LLR

Professor

Dr. Gulde

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Chemistry 102

REVIEW QUESTIONS

TEST 3

1. Calculate the pH of a solution containing 0.100 M sodium formate (NaHCO2)and 0.180 M

formic acid (HCHO2). (Ka formic acid = 1.8x10–4)

2. A buffer is prepared by adding 5.00 g of NH3 and 20.0 g of NH4Cl to enough water to form

2.50 L of solution.

a) Calculate the pH of this buffer.

b) Calculate the pH of the solution when 0.200 g of NaOH is added to 1.50 L of this buffer.

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Chemistry 102

3. Will Mn(OH)2 precipitate from solution if the pH of a 0.050 M solution of MnCl2 is

adjusted to 8.00? (Ksp of manganese hydroxide is 1.9x10–13)

4. Calculate the [Cu2+] in 1.0 L of a solution that contains 1.0x10–3 mole of Cu(NO3)2 and

that is 0.10 M in NH3. (Kf for Cu(NH3)42+ = 5.0x1012)

5. Consider the titration of 25.0 mL of 0.0800 M H2CO3 with 0.100 M KOH. Calculate the

pH of the solution after (a) addition of 10.0 mL of base, and (b) at equivalence point.

(Ka for carbonic acid = 4.3x10–7)

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Document Summary

Test 3: calculate the ph of a solution containing 0. 100 m sodium formate (nahco2)and 0. 180 m formic acid (hcho2). (ka formic acid = 1. 8x10. 4: a buffer is prepared by adding 5. 00 g of nh3 and 20. 0 g of nh4cl to enough water to form. 2. 50 l of solution: calculate the ph of this buffer, calculate the ph of the solution when 0. 200 g of naoh is added to 1. 50 l of this buffer. Chemistry 102: will mn(oh)2 precipitate from solution if the ph of a 0. 050 m solution of mncl2 is adjusted to 8. 00? (ksp of manganese hydroxide is 1. 9x10. 13: calculate the [cu2+] in 1. 0 l of a solution that contains 1. 0x10 that is 0. 10 m in nh3. (kf for cu(nh3)4. 3 mole of cu(no3)2 and: consider the titration of 25. 0 ml of 0. 0800 m h2co3 with 0. 100 m koh.

Related Questions

1. A buffer is made by mixing 1.250 mol formic acid, HCOOH, with 0.750 mol sodium formate, HCOONa in enough water for a total volume of 1.00L. The Ka of formic acid is 1.87 x 10-4

a. What is the pH of this buffer?

b. If 0.075 mol HCl is added to this buffer, what is the pH of the resulting buffer? (assume no volume changes)

c. If 0.055 mol NaOH is added to this buffer, what is the pH of the resulting buffer? (Assume no volume changes)

2. A buffer solution contains 0.120 moles of propionic acid, HC3H5O2, and 0.100 moles of sodium propionate, NaC3H5O2, in 1.00 L of solution. Ka = 1.32 x 10-5 a. What is the pH of this buffer?

b. What is the pH of this buffer after the addition of 0.010 mol of NaOH?

c. Beginning with the initial buffer from part a, what is the pH of the buffer after the addition of 0.010 mol of HBr?

Calculate the pH of the following buffer solutions: (anwers given in bold)

1) A solution obtained by adding 30 mL of 0.20 M HOAc to 25 mL of 0.10 M NaOH. R = 4.61

2) A solution that is 0.050 M in formic acid and 0.10 M in sodium form. R = 4.0

3) A buffer solution of acetic acid-sodium acetate of pH 5.00 (initial) whose concentration in NaOAc is 0.100 M, after the addition of 10 mL of 0.1 M NaOH to 100 mL of the buffer. R = 5.12

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make 1.00 L of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. (a) Calculate the pH of each buffer. (b) Which buffer will have the greater buffer capacity? (c) Calculate the change in pH for each buffer upon the addition of 1.0 mL of 1.00 M HCl. (d) Calculate the change in pH for each buffer upon the addition of 10 mL of 1.00 M HCl.

b) The buffer made from 1.00 mol of each component has a greater buffer capacity, as it can accept greater amount of strong acid or strong base without changing a pH value significantly. The greater is the concentration of the buffer components, the greater is the buffer capacity.

c) The answer in part b) will be confirmed in this part. Adding 1.00 mL of 1.0 M HCl we are adding:

The HCl added reacts with HCOONa:

HCOONa + HCl HCOOH + NaCl

The number of moles of buffer components after addition of HCl:

The pH change is negligible.

The pH of a buffer decreases by 0.09 pH units.

c) Adding 10.0 mL of 1.0 M HCl we are adding:

The HCl added reacts with HCOONa:

HCOONa + HCl HCOOH + NaCl

The number of moles of buffer components after addition of HCl:

The pH changes by 0.01 units.

There is no basic component of a buffer anymore, so the pH is calculated as a pH of a weak acid: