please I need help desperately 1. A student ttrated 25.0 mlL of a buffer with 0.

Question

please I need help desperately

1. A student ttrated 25.0 mlL of a buffer with 0.126 M HCI. The observed titration curve is shown below. What is the concentration of the weak base in the bufier? Thoroughly document your logic including equivalence point and mole conversion. (Hint this is not a dilution (do not use Buffer pH with Added HO 13 2. The student then titrated 25.0mL of the same buffer with 0.115 M NaOH. The observed titration curve is shown below. What is the concentration of the weak acid in the buffer? Thoroughly document your logic including equivalence point and mole conversion. (Hint: this is not a dilution (do not use MV1 M2V2) this is a titration) Buffer pH With Added NaOH 13.00 11.00 7.00 E.00 5.00 3. What is the buffer strength (i.e. the total of [HAl and IAD? 4. If the ph of the buftter (before it is urated) is 5.70, what is the pka of the weak acid? Hint solve the Henderson-Hasselbalch equation for pka

Explanation / Answer

Titration of buffer

1. From the given titration curve, look at the vertical line forming in the graph at around 20 ml of HCl addition.

This is your equivalence point

moles of HCl added to reach equivalence point = molarity x volume

                                                                            = 0.126 M x 20 ml = 2.52 mmol

moles of acid reacting = moles of weak base present in buffer = 2.52 mmol

So,

concentration of weak base in the buffer = 2.52 mmol/25 ml = 0.101 M

2. Similarly, looking at the graph, the vertical line is forming at around 11 ml of NaOH added

This is your equivalence point

moles of NaOH added to reach equivalence point = molarity x volume

                                                                                = 0.115 x 11 ml = 1.265 mmol

moles of acid reacting = moles of weak acid present in buffer = 1.265 mmol

So,

concentration of weak acid in the buffer = 1.265 mmol/25 ml = 0.0506 M

3. buffer strength = [HA] + [A-]

                            = 0.0506 + 0.101 = 0.1516 M

4. Using Hendersen-Hasselbalck equation,

pH = pKa + log([A-]/[HA])

with,

pH = 5.70

[HA] = 0.0506 M

[A-] = 0.101 M

we get,

pKa = pH - log([A-]/[HA])

       = 5.70 + log(0.101/0.0506) = 6.00

So the pKa of buffer is approximately around 6.00.

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