Question

Question 2:

A) Calculate the pH of the buffer that results from mixing 56.1 mL of a 0.406 M solution of HCHO2 and 11.9 mL of a 0.606 M solution of NaCHO2. The Ka value for HCHO2 is 1.8×10−4

B) Calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

  300.0 mL of a buffer solution that is 0.225 M in HCHO2 and 0.280 M in KCHO2

C) Calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

300.0 mL of a buffer solution that is 0.280 M in CH3CH2NH2 and 0.260 M in CH3CH2NH3Cl

D)

Which acid is the best choice to create a buffer with pH= 1.83?

Which acid is the best choice to create a buffer with 1.83?

chlorous acid (HClO2),pKa=1.95

hypochlorous acid (HClO),pKa=7.54

nitrous acid (HNO2),pKa=3.34

acetic acid (HCH3CO2),pKa=4.76

Answer 1

A)
pH of acidic buffer = pka + log(HcooNa/HCOOH)

pka of HCOOH = -logKa = -log(1.8*10^-4) = 3.74

no of mol of HCOOH = 56.1*0.406 = 22.78 mol

No of mol of HCOONa =   11.9*0.606 = 7.2 mmol

pH = 3.74+log(7.2/22.78)

    = 3.24

B) initial pH

pH of acidic buffer = pka + log(HcooNa/HCOOH)

pka of HCOOH = -logKa = -log(1.8*10^-4) = 3.74

concentration of HCOOH = 0.225 M

Concentration of HCOONa = 0.28 M

pH = 3.74+log(0.28/0.225)

    = 3.83

final pH

pH of acidic buffer = pka + log(HcooK+NaOH/HCOOH-NaOH)

pka of HCOOH = -logKa = -log(1.8*10^-4) = 3.74

concentration of HCOOH = 0.225 M

Concentration of HCOOk = 0.28 M

Concentration of NaOH added = n/v = 0.01/0.3 = 0.033 M

pH = 3.74+log((0.28+0.033)/(0.225-0.033))

pH = 3.95

C)

Basic buffer, initial pH

pOH = pkb+log(CH3CH3NH3Cl/CH3CH3NH2)

pkb of CH3CH3NH2 = -log(Kb) = -log(4.3*10^-4) = 3.36

concentration of CH3CH3NH2 = 0.28 M

Concentration ofCH3CH3NH3cl = 0.26 M

pOH = 3.36+log(0.26/0.28)

    = 3.33

pH = 14-3.33 = 10.67

final pH

pOH = pkb+log(CH3CH3NH3Cl- NaOH/CH3CH3NH2+NaOH)

pkb of CH3CH3NH2 = -log(Kb) = -log(4.3*10^-4) = 3.36

concentration of CH3CH3NH2 = 0.28 M

Concentration ofCH3CH3NH3cl = 0.26 M

Concentration of NaOH added = n/v = 0.01/0.3 = 0.033 M

pOH = 3.36+log((0.26-0.033)/(0.28+0.033))

    = 3.22

pH = 14-3.22 = 10.78

D) answer: chlorous acid (HClO2),pKa=1.95