Question

each solut 39. Use the Henderson-Hasselbalch equal each solution a. a solution that is 0.135 M in HCIO and 0.155 M in KCIO b. a solution that contains 1.05% CH3NH, by mass and 1.10% CH NH;Br by mass c. a solution that contains 10.0 g of HC H2O2 and 10.0 g of NaC,H,O, in 150.0 mL of solution 40. Use the Henderson-Hasselbalch equation to calculate the pH of each solution. a. a solution that is 0.145 M in propanoic acid and 0.115 M in potassium propanoate b. a solution that contains 0.785% C3H5N by mass and 0.985% C3H NHCl by mass c. a solution that contains 15.0 g of HF and 25.0 g of NaF in 125 mL of solution 41. Calculate the pH of the solution that results from each mixture. a. 50.0 mL of 0.15 M HCHO2 with 75.0 mL of 0.13 M NaCHO2 b. 125.0 mL of 0.10 M NHz with 250.0 mL of 0.10 M NH C1 42. Calculate the nH of the solution th

answer number 40, a and c with all work shown,
the ka value for HF us 3.5 x 10^-4

Answer 1

Answey 40. (a). Buffer is made up of weak acid, propapcic acid and its salt potassium propanoate. given, concentration og propancic acid = [CH3 CH2 2001)=0-145M. and concentration of potassium propanoate = [CH3 CH2cook] = 0.115 pka g propancic acid = 4.87 aceonding to Henderson - Hasselbalch equation, - pH = pka + log [A] where, [HA] = concentration of weak acid. [AO] = concentration of salt of weak aciol. .: pH= 4.87ilog 0.115 - pH= 4.87+ boso (-0.1006). .; PH = 4-8'7- 0.1006. or (pH = 4.77. THA] 0.145

40.(0) We trou moiarity = 030.9 moles o solute. 40.(C) We know motary Volume of solution (1) 15 now, in a given solution, mass og HF = 15.09. molat mass of HF = 20 g/mol. ionumber of males of HF = _given mass Molar mass - 0.75 moles. and volume of seletion = 125 mL :: CHF] = 120.01 moles HF 0.125L. Volume g sopot = 0.75 mot alom. and similarly, we have, mass of NaF = 259. molar mass of NaF =°23 + 19 = 429/mol. .: no.of moles of NaF = 25/42 =0.59 mot. .: [NaF] = [NFO]= 0.59 = 4.72. M. given. Ka fon HF = 3.5 X10-4 hence, pka = -log (3-5 X10-4) [pKQ = -log ka] .. = 3.45. using Henderson, Hasselbalth equation, pH = pka + log_[Fe] Y THET :: PH = 3.45+ log 4•72 :,: pH = 3.45 -0.10 PH = 3.35 0.125 =