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Part A: Calculate the pH of a solution that is 0.250 M in sodium formate (HCOONa)...

Part A:

Calculate the pH of a solution that is 0.250 M in sodium formate (HCOONa) and 0.110 M in formic acid (HCOOH).

Part B:

Calculate the pH of a solution that is 0.520 M in pyridine (C5H5N) and 0.470 M in pyridinium chloride (C5H5NHCl).

Part C:

Calculate the pH of a solution that is made by combining 55 mL of 0.040 M hydrofluoric acid with 125 mL of 0.120 M sodium fluoride.

Solutions

Expert Solution

Part A:

Calculate the pH of a solution that is 0.250 M in sodium formate (HCOONa) and 0.110 M in formic acid (HCOOH).

Solution :- weak acid + conjugate base of weak acid makes the buffer solution therefore using the Henderson equation we can calculate the pH

Pka of the formic acid is = 3.75

Using the Henderson equation we can calculate the pH of the solution

pH= pka + log [base]/[acid]

pH= 3.75 + log[0.250]/[0.110]

pH= 4.11

Part B:

Calculate the pH of a solution that is 0.520 M in pyridine (C5H5N) and 0.470 M in pyridinium chloride (C5H5NHCl).

Solution :-

pKb of pyridine = 8.77

pka of puridine = 14 – 8.77 = 5.23

pH= pka + log [base]/[acid]

pH= 5.23 + log [0.520]/[0.470]

pH= 5.27

Part C:

Calculate the pH of a solution that is made by combining 55 mL of 0.040 M hydrofluoric acid with 125 mL of 0.120 M sodium fluoride.

Solution :-

Total volume of solution = 55ml + 125 ml = 180 ml

New molarity of the hydrofluoric acid at the total volume is calculated as follows

M1V1=M2V2

M2= M1V1/V2

New molarity of [HF] M2 = 0.040 M * 55 ml / 180 ml = 0.0122 M

New molarity of the [NaF] M2 = 0.120 M* 125 ml / 180 ml = 0.0833 M

Now lets calculate the pH

Pka of HF acid = 3.18

pH= pka + log[base]/[acid]

pH= 3.18 + log[0.0833]/[0.0122]

pH= 4.01

queen_honey_blossom answered 1 year ago

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