A 50.0 mL solution of 0.105 M KOH is titrated with 0.210 M HCL. calculate the pH

Question

A 50.0 mL solution of 0.105 M KOH is titrated with 0.210 M HCL. calculate the pH of the solution after the addition of the following of HCL.

of 10 f0 10 t citrated with 0:210 M HCI. Calculate the pH of the solution after th addition of the following amounts of HCl. a) 0.00 mL HCI e) 24.0 mL HCI Number Number pH- b) 5.00 mL HCI f) 25.0 mL HCI Number Number C) 12.5 mL HCI g) 26.0 mL Hc Number umber PH d) 18.0 mL HCI h) 30.0 mL HCI Number Number PH Previou Give tip&ew lichaibril G) Check Answer Next

Explanation / Answer

Number of moles of KOH = 0.105*50 = 5.25 mmol

pOH = -log[OH-]

pOH = -log(5.25/50)

pOH = 0.9788

pH = 14 - 0.9788 = 13.0212

b) number of moles of HCl = 5 * 0.21 = 1.05 mmol

number of moles remainig = 5.25 - 1.05 = 4.2 mmol

pOH = -log(4.2/(50+5))

pOH = 1.1171

pH = 14 - 1.1171

pH = 12.889

c) number of moles of HCl = 12.5 * 0.21 = 2.625 mmol

number of moles remainig = 5.25 - 2.625 = 2.625 mmol

pOH = -log(2.625/(50+12.5))

pOH = 1.377

pH = 14 - 1.377 = 12.623

d) number of moles of HCl = 18 * 0.21 = 3.78 mmol

number of moles remainig = 5.25 - 3.78 = 1.47 mmol

pOH = -log(1.47/(50+18))

pOH = 1.6652

pH = 14 - 1.6652 = 12.3348

e) number of moles of HCl = 24 * 0.21 = 5.04 mmol

number of moles remainig = 5.25 - 5.04 = 0.21 mmol

pOH = -log(0.21/(50+24))

pOH = 2.547

pH = 14 - 2.547 = 11.453

f) number of moles of HCl = 25 * 0.21 = 5.25 mmol

number of moles remainig = 5.25 - 5.25 = 0.0 mmol

this is the equivalence point so pH = 7

g) number of moles of HCl = 26 * 0.21 = 5.46 mmol

number of moles remainig = 5.46 - 5.25 = 0.21 mmol

pH = -log(0.21/(50+26))

pH = 2.5586

h) number of moles of HCl = 30 * 0.21 = 6.3 mmol

number of moles remainig = 6.3 - 5.25 = 1.05 mmol

pH = -log(1.05/(50+30))

pH = 1.882

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