e-CHE 168-Spring18-MCGREGOR Activities and Due Dates Topic 7 Skill Practice Víde

Question

e-CHE 168-Spring18-MCGREGOR Activities and Due Dates Topic 7 Skill Practice Vídeo A G1 2018 1 1:55 PM 2/3 () 3/2/2018 03:35 PM Print CalculatorPeriodic Table n 3 of 3 pH of the solution after the Ma A 50.0 mL solution of 0.136 M KOH is titrated with 0272 M HCI. Calculate the addition of the following amounts of HCI a) 0.00 mL HCI e) 24.0 mL HCI Number Number b) 7.00 mL HC 25.0 mL HCI Number Number c) 12.5 mL HCI 9) 26.0 mL HCI Number Number d) 19.0 mL HC h) 31.0 mL HCI Number Number O Previous Give Up & View Soluton D Check Answer Next ] Exit Hint

Explanation / Answer

millimoles of KOH = 50 x 0.136 = 6.8

a) before addition of any HCl

KOH = 0.136

[OH-] = 0.136 M

pOH = -log[OH-] = -log (0.136) = 0.866

pH +pOH = 14

pH = 13.13

b) 7.00 ml HCl added

millimoles of HCl = 7 x 0.272 = 1.904

millimoles of base > millimoles of acid

so [OH-] = (base millimoles - acid millimoles )/ total volume

               = (6.8 - 1.904) / (50 + 7)

              = 0.08589 M

pOH = 1.066

pH = 12.93

c) 12.5 ml HI added

millimoles of HCl = 12.5 x 0.272 = 3.4

[OH-] = 6.8 - 3.4 / 50 + 12.5 = 0.0544 M

pH = 12.74

d) 19 mL added

[OH-] = 0.02365

pH = 12.37

e)

[OH-] = 3.676 x 10^-3 M

pH = 11.56

f)

pH = 7.00

g

pH = 1.86

h)

pH = 1.70

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