Part A For 540.0 mL of pure water, calculate the initial pH and the final pH aft
ID: 509310 • Letter: P
Question
Part A For 540.0 mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of HCl. Express your answers using two decimal places separated by a comma.
pHinitial, pHfinal =
Part B For 540.0 mL of a buffer solution that is 0.125 M in HC2H3O2 and 0.110 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.010 mol of HCl. Express your answers using two decimal places separated by a comma.
pHinitial, pHfinal=
Part C For 540.0 mL of a buffer solution that is 0.155 M in CH3CH2NH2 and 0.140 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of HCl. Express your answers using two decimal places separated by a comma.
pHinitial, pHfina=l
Explanation / Answer
Volume of solution = 540mL
initially it was pure water , its pH = 7.00
amount of Hcl = 0.010mol
Thus molarity of solution = 0.010mole/0.540L
= 0.01851 M
Thus the [H+] = 0.01851
pH = -log [H+]
= 1.732
Part B
volume of buffer = 540mL
[acid] = 0.125M
[conjugate base]= 0.125M
The pH of the buffer is given by Hendersen equation
pH = pKa + log [conjugate base]/[acid]
= 4.76 + log 0.125/0.125
= 4.76
after adding 0.010 mol HCl
HA ------------> A- + HCl
0.125 x540 0.125x540 0 initial mmoles
=67.5 = 67.5 0
- - 0.010mol=10mmol
77.5 57.5 0
Thus the pH of buffer after adding HCl
pH = 4.76 + log 57.5/77.5
= 4.63
Part C
pKb of ethylamine = 3.3
[base] =0.155M
[conjugate acid] = 0.140M
pH of basic buffer = 14-[pOH +log [conjugate acid]/[base]}
=14 - [3.3 + log 0.140/0.155]
= 10.744
B +HCl -------------------> BH+ + Cl-
540x0.155 0 540x0.14 - initial mmoles
=83.7 =75.6
10 - change
73.7 0 85.6 after reaction
Thus pH after addition foHCl is
pH = 14- {pOH + log [conjugate acid]/[base]}
= 14 - { 3.3 + log 85.6/73.7}
=10.6349