Youve Been Asked By Your Research Advisor To

Question

Youve been asked by your research advisor to prepare 250.00 mL of a 0.020 M buffer. The target pH…

appropriate sig figs!!! please

Youve been asked by your research advisor to prepare 250.00 mL of a 0.020 M buffer. The target pH is 5.60. The components are C3Hs02 HC3Hs02 and the Ka of the weak acid is 1.3 x 105. Part A what volume (in mL) of a 0.050 M C3H502-solution is taquired to prepare the desired buffer? ml Submit Request Answer Part B What volume (in mL) of a 0.050 M HC Hs02 solution is required to prepare the desired buffer? ml

Solutions

Expert Solution

Part A : Volume of
C3H5O2– = 83.8
mL

Part B : Volume of HC3H5O2 =
16.2 mL

Explanation

According to Henderson Hasselbalch equation

pH = pKa + log([conjugate base] / [weak acid])

pH = pKa +
log([C3H5O2–] /
[HC3H5O2])

5.60 = 4.886 +
log([C3H5O2–] /
[HC3H5O2])

log([C3H5O2–] /
[HC3H5O2]) = 5.60 – 4.886

log([C3H5O2–] /
[HC3H5O2]) = 0.714

[C3H5O2–] /
[HC3H5O2] =
10(0.714)

[C3H5O2–] /
[HC3H5O2] =
5.1754    …(1)

Also, [C3H5O2–] +
[HC3H5O2] = 0.020
M    …(2)

Solving equation (1) and (2)

[C3H5O2–] = 0.01676
M

[HC3H5O2] = 0.00324 M


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