A 100.0-mL buffer solution is 0.180 molL1 in HClOand 0.135 molL1 in NaClO. K a(H
ID: 229164 • Letter: A
Question
A 100.0-mL buffer solution is 0.180 molL1 in HClOand 0.135 molL1 in NaClO.
Ka(HClO)=4.0×108
Part A
What is the initial pH of this solution?
Express your answer using two decimal places.
Part B
What is the pH after addition of 115.0 mg of HBr?
Express your answer using two decimal places.
Part C
What is the pH after addition of 85.0 mg of NaOH?
Express your answer using two decimal places.
A 100.0-mL buffer solution is 0.180 molL1 in HClOand 0.135 molL1 in NaClO.
Ka(HClO)=4.0×108
Part A
What is the initial pH of this solution?
Express your answer using two decimal places.
Part B
What is the pH after addition of 115.0 mg of HBr?
Express your answer using two decimal places.
Part C
What is the pH after addition of 85.0 mg of NaOH?
Express your answer using two decimal places.
Explanation / Answer
no of moles of HClO = molarity * volume in L
= 0.18*0.1 = 0.018 moles
no of moles of NaClO = molarity * volume in L
= 0.135*0.1 = 0.0135 moles
Pka = -logKa
= -log4*10^-8
= 7.3979
PH = PKa + log[NaClO]/[HClO]
= 7.3979 + log0.0135/0.018
= 7.3979-0.1249 = 7.273
part-B
no of moles of HBr = W/G.M.Wt
= 0.115/81 = 0.00142moles
no of moles of HClO after addition of 0.00142 moles of HBr = 0.018 +0.00142 = 0.01942moles
no of moles of NaClO after addition 0.00142 moles of HBr = 0.0135-0.00142 = 0.01208 moles
PH = PKa + log[NaClO]/[HClO]
= 7.3979+ log0.01208/0.01942
= 7.3979-0.2061
= 7.1918
part-C
no of moles of NaOH = W/G.M.Wt
= 0.085/40 = 0.002125 moles
no of moles of HClO after addition of 0.002125 moles of NaOH = 0.018 -0.002125 = 0.015875moles
no of moles of NaClO after addition 0.002125 moles of NaOH = 0.0135+0.002125 = 0.015625 moles
PH = PKa + log[NaClO]/[HClO]
= 7.3979+ log0.015875/0.015625
= 7.3979 + 0.006894 = 7.4048
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