100.0 mL buffer solution contains 0.500 M HClO (Ka= 2.9x10^-8) and 0.450 KClO. W
ID: 1044611 • Letter: 1
Question
100.0 mL buffer solution contains 0.500 M HClO (Ka= 2.9x10^-8) and 0.450 KClO. What is the initial pH of the buffer solution? What is the pH of the solution after the addition of 0.0050 moles of HBr? What is the pH of the solution after the addition of 0.010 moles of KOH? 100.0 mL buffer solution contains 0.500 M HClO (Ka= 2.9x10^-8) and 0.450 KClO. What is the initial pH of the buffer solution? What is the pH of the solution after the addition of 0.0050 moles of HBr? What is the pH of the solution after the addition of 0.010 moles of KOH? What is the initial pH of the buffer solution? What is the pH of the solution after the addition of 0.0050 moles of HBr? What is the pH of the solution after the addition of 0.010 moles of KOH? What is the pH of the solution after the addition of 0.0050 moles of HBr? What is the pH of the solution after the addition of 0.010 moles of KOH?Explanation / Answer
no of moles of HClO = molarity * volume in L
= 0.5*0.1 = 0.05moles
no of moles of KClO = molarity * volume in L
= 0.45*0.1 = 0.045mole
PKa = -logKa
= -log2.9*10^-8
= 7.5376
PH = Pka + log[KClO]/[HClO]
= 7.5376 + log0.045/0.05
= 7.5376 -0.04575 = 7.4918
no of moles of KClO after addition of 0.005 moles of HBr =0.045-0.005 = 0.04moles
no of moles of HClO after addition of 0.005moles of HBr = 0.05+0.005 = 0.055moles
PH = PKa + log[KClO]/[HClO]
= 7.5376 + log0.04/0.055
= 7.5376 -0.1383 = 7.3993
no of moles of KClO after addition of 0.01 moles of KOH =0.045+0.01 = 0.055moles
no of moles of HClO after addition of 0.01moles of HBr = 0.05-0.01 = 0.04moles
PH = PKa + log[KClO]/[HClO]
= 7.5376 + log0.055/0.04
= 7.5376 +0.1383 = 7.6759
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