For each of the following solutions, calculate the initial pH and the final pH a
ID: 1004419 • Letter: F
Question
For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.Part A For 260.0 mL of pure water, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.Part B For 260.0 mL of a buffer solution that is 0.200 M in HCHO2 and 0.310 M in KCHO2, calculate the initial pH and the final pH after adding 0.005 mol of NaOH. For 260.0 mL of a buffer solution that is 0.320 M in CH3CH2NH2 and 0.295 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.005 mol of NaOH
Explanation / Answer
Pure water PH value is 7( initial PH =7)
no of moles of NaOH = 0.005moles
molarity of NaOH = no of moles of NaOH /volume in L
= 0.005/0.26 = 0.0192M
NaOH is strong base
NaOH ------> Na+ + OH-
0.0192M 0.0192M
[OH-] = 0.0192M
POH = -log[OH-]
= -log0.0192
= 1.717
PH = 14-POH
= 14-1.717 = 12.283( final PH value)
B. pka of HCOOH is 3.74
no of moles of HCHO2 = molarity * volume in L
= 0.2*0.26 = 0.052moles
no of moles of KCHO2 = 0.31*0.26
= 0.0806moles
PH = PKa + log[KCHO2]/[HCHO2]
= 3.74 + log0.0806/0.052
= 3.74+0.19 = 3.93(initial PH value)
by the addition of 0.005moles of NaOH
no of moles of HCHO2 = 0.052-0.005 = 0.047 moles
no of moles of KCHO2 = 0.0806+0.005 = 0.0856 moles
PH = PKa + log[KCHO2]/[HCHO2]
= 3.74+ log0.0856/0.047
= 3.74+ 0.26 = 4 ( final PH value)
C. no of moles of CH3CH2NH2 = molarity* volume in L
= 0.32*0.26 = 0.0832 moles
no of moles of CH3CH2NH3Cl = 0.295*0.26 = 0.0767moles
POH = PKb + log[CH3CH2NH3Cl]/[ CH3CH2NH2]
= 3.37 + log0.0767/0.0832
= 3.37-0.0353 = 3.3347
PH = 14-POH
= 14-3.3347 = 10.6653 (initial PH value)
by the addition of 0.005 moles of NaOH
no of moles of CH3CH2NH3Cl = 0.0767- 0.005 = 0.0717 moles
no of moles of CH3CH2NH2 = 0.0832 +0.005 = 0.0882 moles
POH = PKb + log[CH3CH2NH3Cl ]/[CH3CH2NH2]
= 3.37+ log0.0717/0.0882
= 3.37-0.0899 = 3.2801
PH = 14-POH
= 14-3.2801 = 10.7199 ( final PH)
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