Question

Part A

Calculate the pH of 0.100 L of a buffer solution that is 0.27 M in HF and 0.47 M in NaF.

Express your answer using three significant figures.

pH =

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Part B

Calculate pH of the solution on addition of the following.
0.004 mol of HNO3

Express your answer using three significant figures.

pH =

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Part C

Calculate pH of the solution on addition of the following.
0.005 mol of KOH

Express your answer using three significant figures.

pH =

Answer 1

A)

Ka of HF = 6.6*10^-4

pKa = - log (Ka)

= - log(6.6*10^-4)

= 3.1805

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.1805+ log {0.47/0.27}

= 3.42

Answer: 3.42

B)

mol of HNO3 added = 0.004 mol

F- will react with H+ to form HF

Before Reaction:

mol of F- = 0.47 M *0.1 L

mol of F- = 0.047 mol

mol of HF = 0.27 M *0.1 L

mol of HF = 0.027 mol

after reaction,

mol of F- = mol present initially - mol added

mol of F- = (0.047 - 0.004) mol

mol of F- = 0.043 mol

mol of HF = mol present initially + mol added

mol of HF = (0.027 + 0.004) mol

mol of HF = 0.031 mol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.1805+ log {0.043/0.031}

= 3.32

Answer: 3.32

c)

mol of KOH added = 0.005 mol

HF will react with OH- to form F-

Before Reaction:

mol of F- = 0.47 M *0.1 L

mol of F- = 0.047 mol

mol of HF = 0.27 M *0.1 L

mol of HF = 0.027 mol

after reaction,

mol of F- = mol present initially + mol added

mol of F- = (0.047 + 0.005) mol

mol of F- = 0.052 mol

mol of HF = mol present initially - mol added

mol of HF = (0.027 - 0.005) mol

mol of HF = 0.022 mol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.1805+ log {0.052/0.022}

= 3.55

Answer:3.55