pH of a Buffered Solution Please answer 3 questions. Using tabulated K a and K b

Question

pH of a Buffered Solution

Please answer 3 questions.

Using tabulated Ka and Kb values, calculate the pH of the following solutions.
Answer to 2 decimal places.

A solution containing 3.5×10-1 M HCOOH and 3.6×10-1 M HCOONa.

750.0 mL ofsolution containing 9.0 g of HF and 35.0 g of NaF.

A solution made by mixing 100.00 mL of 0.950 M HNO2 with 65.00 mL of 0.900 M NaOH

Ka & Kb values may be found in Zumdahl, Chemical Principles 8th ed. Appendix 5 Tables A5.1, A5.3. pH of a Buffered Solution Using tabulated Ka and Kb values, calculate the pH of the following solutions. Answer to 2 decimal places. Ka & Kb Values may be found in Zumdahl, Chemical Principles 8th ed. Appendix 5 Tables A5.1, A5.3. A solution containing 3.5 x 10-1 M HCOOH and 3.6x 10-1 M HCOONa. ipts Submit Answer Tries 0/5 750.0 mL of solution containing 9.0 g of HF and 35.0 g of NaF. 1pts Submit Answer Tries 0/5 A solution made by mixing 100.00 mL of 0.950 M HNo2 with 65.00 mL of 0.900 M NaOH. 1pts Submit Answer Tries 0/5

Explanation / Answer

a)

this is a buffer so

pH = pKa + log(HCOONa / HCOOH)

pKa = 3.75 for formic acid

so

pH= 3.75 + log((3.6*10^-1)/(3.5*10^-1) )

pH = 3.76

b)

mol of HF = mass/MW = 9/20 = 0.45

mol of NaF = mass/MW = 35/41.98817 = 0.8335

pH = pKa + lof(NaF/HF)

pKa for HF = 3.14

pH = 3.14 + log(0.8335/0.45)

pH = 3.407

c)

mmol of HNO2 = MV = 0.95*100 = 95

mmol of NaOH = MV = 0.9*65 = 58.5

after reaction

mmol of HNO2 left = 95- 58.5 = 36.5

mmol of NO2- formed = 58.5

pH = pKa + log(NO2-/HNO2)

pH = 3.32 + log(58.5 /36.5)

pH = 3.524

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