Calculate the pH of the solution that results from each of the following mixture

Question

Calculate the pH of the solution that results from each of the following mixtures.

Part A

160.0 mL of 0.23 M HF with 220.0 mL of 0.31 M NaF

Express your answer using two decimal places.

Part B

185.0 mL of 0.11 M C2H5NH2 with 285.0 mL of 0.22 M C2H5NH3Cl

Express your answer using two decimal places.

Calculate the pH of the solution that results from each of the following mixtures.

Part A

160.0 mL of 0.23 M HF with 220.0 mL of 0.31 M NaF

Express your answer using two decimal places.

pH =

Part B

185.0 mL of 0.11 M C2H5NH2 with 285.0 mL of 0.22 M C2H5NH3Cl

Express your answer using two decimal places.

pH =

Explanation / Answer

Part A

160.0 mL of 0.23 M HF with 220.0 mL of 0.31 M NaF

Here the total volume = 160.0+220.0 = 380.0 mL
Now molarity of HF=

160*0.23/380= 0.097 M

Molarity of NaF=

220*0.31/380 = 0.179 M

The pKa of HF = 3.17

Calculate the pH using the Henderson - Hasselbalch equation as follows:
pH = pKa + log ([salt]/[acid])
pH =- 3.17 + log ( 0.179/0.097)
pH = 3.17 + log 1.85
pH = 3.17 + 0.267
pH = 3.44

Part B

185.0 mL of 0.11 M C2H5NH2 with 285.0 mL of 0.22 M C2H5NH3Cl

Here the total volume = 185.0+285.0 =470.0 mL
Now molarity of C2H5NH2=

185*0.11/470= 0.0433 M

Molarity of C2H5NH3Cl =

285*0.22/470 = 0.1334 M

pKa for ethylamine or C2H5NH2 is 10.7

C2H5NH2 is a base, pKb = 3.3

The Henderson-Hasselbalch equation for basic buffer is

pOH = pKb + log[C2H5NH3+]/[C2H5NH2]

pOH = pKb + log 0.1334/ 0.0433
pOH = 3.3 + 0.489
pOH = 3.72 = 3.789

pH = 14-Poh

pH=14-3.789

=10.21

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