6. Calculate the pH of the 0.20 M NH 3 /0.20 M NH 4 Cl buffer. What is the pH of

Question

6.      Calculate the pH of the 0.20 M NH3/0.20 M NH4Cl buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 65.0 mL of the buffer?

7.      A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid.

8.      In a titration experiment, 12.5 mL of 0.500 M H2SO4 neutralize 50.0 mL of NaOH. What is the concentration of the NaOH solution?

9.      A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87. What is the Ka of the unknown acid?    

10. Calculate the pH at the equivalence point for the following titration: 0.20 M HCl versus 0.20 M methylamine (CH3NH3; Kb = 4.4

Explanation / Answer

6) You need to use the Henderson Hasselbalch equation pH=pKa + log([base]/[acid])

The addition of HCl changes the amounts of NH3 and NH4+

.1 M HCl * .015 L = .0015 moles HCl will react with moles of NH3.

Finding moles NH3: .065 L * .2 M NH3 = .013 moles NH3

When H+ ions (from the HCl) react with NH3, it will decrease NH3, but will increase NH4+ concentration.

pH = 9.25 (pKa of NH4+) + log ( (.013 moles NH3 - .0015 moles H+)/( .013 moles NH4+ + .0015 moles NH4+) )

= 9.149 pH
Is this answer realistic? Yes, a buffer solution resists dramatic changes in pH from the addition of both acids and bases. We did add an acid, so the pH should have dropped a little from the pKa value which is the pH of soln. when the concentrations are equal. The concentrations were equal at the beginning, so the pH was 9.25 initially, dropping to 9.15 after the addition of HCl.

7)moles KOH = 0.08133 M x 0.0164 L=0.00133 = moles acid

molar mass = 0.2688 g/ 0.00133=201.5 g/mol

8)H2SO4 has two protons so the is 2X 0.500 moles/L X .0125 L = 0.0125 moles of acid. 0.0125moles / 0.050L = 0.25M NaOH.

9)find moles,

0.0184 litres @ 0.0633 mol / litre = 0.001165 moles NaOH

By the equation 1 NaOH & 1 HA --> NaA & H2O
0.001165 moles NaOH reacts with 0.001165 moles Ha

a) Calculate the molar mass of the acid
0.1276 grams / 0.001165 moles =
your answer
109.6 grams / mole
======================================...

find how much acid is lost by 10 ml's into the titration:
0.001165 mol total @ 10 ml / 18.4 ml =
0.000633 moles of acid have been neutralized by pH of 5.87

those 0.000633 moles of acid have been converted into 0.000633 moles of anion "A-"

0.001165 mol total - those 0.000633 moles of acid = have 0.000532 moles of acid left


by the henderson-hasselbalch equation:

pH = pKa + [A-] / [HA]

5.87 = pKa + [0.000633 ] / [0.000532]

5.87 = pKa + 1.19

pKa = 4.68

Ka = 2.1 e -5

10)assume 1 L of CH3NH2

moles CH3NH2 = 0.020
moles H+ needed = 0.020
Volume HCl = 0.020 mol/ 0.020 M = 1 L
total volume = 2 L
CH3NH2 + H+ = CH3NH3+
[CH3NH3+]= 0.020/ 2 = 0.01 M

Ka = Kw/Kb = 2.27 x 10^-11 = x^2 / 0.01-x
x = [H+]= 4.77 x 10^-7 M
pH = 6.32

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