Question

 

1.     What is the ph of a solution made by adding 100g of sodium acetate to 500 ml of water?

2.     What is the PH of a solution that contains 0.1 moles of sodium hydrogen phosphate and 0.1 moles of sodium dihydrogen phosphate?

3.     How many grams of sodium hydrogen phosphate and sodium dihydrogen phosphate are needed to make 200 ml of a 0.15M PH 6.80 buffer?

4.     What is the PH when 10 ml of 0.1 M HCl is added to 100 ml of 0.1 M phosphate buffer that had a starting PH of 7.2?

5.     What volume (mL) of 0.12 M sodium hydroxide was added to 300.00 mL of a 0.15  M phosphate buffer that has a starting pH of 6.70 to change the pH to 7.70   after the addition of this amount of base. ( The pKa is 6.80.)

Answer 1

Solution:

1.

100 g Sodium acetate

500 ml water

Lets first calculate the molarity of the sodium acetate

moles = mass/ molar mass

moles of sodium acetate = 100 g / 82.0343 g per mol 1.22 mol

molarity = moles / volume in liter

Molarity of sodium acetate = 1.22 mol / 0.500 L =2.44 M

When acetate ion dissolved in water then it reacts with water to produce the OH^- ions by following equation

CH3COO^- + H2O   --------- > CH3COOH     + OH^-

2.44 M                                      0                   0

-x                                             +x                  +x

2.44 -x                                      x                    x

Lets write the kb equation

Kb= [CH3COOH][OH^-]/[CH3COO^-]

Kb of sodium acetate is 5.55*10^-10

lets put the values in the formula

5.55*10^-10 = [x][x]/[2.44-x]

since Kb is very small therefore we can neglect the x from the denominator then we get

5.55*10^-10 = [x][x]/[2.44]

5.55*10^-10 * 2.44 = x^2

1.35*10^-9 =x^2

taking square root of both sides we get

3.67*10^-5 = x =[OH-]

now using the [OH-] lets calculate pOH

pOH= -log [OH-]

pOH = -log [3.67*10^-5]

pOH =4.44

Now lets find pH using the pOH

pH + pOH = 14

pH= 14 - pOH

pH= 14 - 4.44

pH= 9.56

Therefore pH of the solution is 9.56