Question

An
analytical chemist is titrating 138.2 mL of a 1.100 M solution of
butanoic acid (HC3H7CO2) with a 0.6000 M solution of KOH. The pKa
of butanoic acid is 4.82. Calculate the pH of the acid solution
after the chemist has added 294.3 mL of the KOH solution to it.
Please round to 2 decimal places.

Answer #1

**use:**

**pKa = -log Ka**

**4.82 = -log Ka**

**Ka = 1.514*10^-5**

**Given:**

**M(HC3H7CO2) = 1.1 M**

**V(HC3H7CO2) = 138.2 mL**

**M(KOH) = 0.6 M**

**V(KOH) = 294.3 mL**

**mol(HC3H7CO2) = M(HC3H7CO2) * V(HC3H7CO2)**

**mol(HC3H7CO2) = 1.1 M * 138.2 mL = 152.02
mmol**

**mol(KOH) = M(KOH) * V(KOH)**

**mol(KOH) = 0.6 M * 294.3 mL = 176.58 mmol**

**We have:**

**mol(HC3H7CO2) = 152.02 mmol**

**mol(KOH) = 176.58 mmol**

**152.02 mmol of both will react**

**excess KOH remaining = 24.56 mmol**

**Volume of Solution = 138.2 + 294.3 = 432.5
mL**

**[OH-] = 24.56 mmol/432.5 mL = 0.0568 M**

**use:**

**pOH = -log [OH-]**

**= -log (5.679*10^-2)**

**= 1.2458**

**use:**

**PH = 14 - pOH**

**= 14 - 1.2458**

**= 12.7542**

**Answer: 12.75**

An
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c.
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e.
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Consider a 2×10−2 M solution of the weak
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You can search for the structure of this compound online,
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Your answers need to be within 5% of the correct answer to be
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Part B: Calculate the concentration of
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Part...

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