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Math  /  Algebra

QuestionAn analytical chemist is titrating 126.3 mL of a 0.8800 M solution of trimethylamine ((CH3)3 N)\left(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\right) with a 0.7800 M solution of HClO4\mathrm{HClO}_{4}. The pKbp K_{b} of trimethylamine is 4.19. Calculate the pH of the base solution after the chemist has added 39.1 mL of the HClO4\mathrm{HClO}_{4} solution to it.
Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HClO4\mathrm{HClO}_{4} solution added. Round your answer to 2 decimal places. pH=\mathrm{pH}= □\square

Studdy Solution

STEP 1

What is this asking? We're figuring out how acidic a trimethylamine solution becomes after we add some perchloric acid to it. Watch out! Don't forget to account for the change in concentration due to the added acid!

STEP 2

1. Calculate moles of base and acid
2. Determine remaining base and conjugate acid
3. Calculate pOH using Henderson-Hasselbalch equation
4. Calculate pH

STEP 3

We **multiply** the volume of the trimethylamine solution by its molarity to find the initial number of moles.
Remember, molarity is moles per liter, so we need to **convert** mL to L first! 126.3 mL=0.1263 L126.3 \text{ mL} = 0.1263 \text{ L}. 0.1263 L⋅0.8800molL=0.111144 mol0.1263 \text{ L} \cdot 0.8800 \frac{\text{mol}}{\text{L}} = \mathbf{0.111144} \text{ mol} So, we start with **0.111144 moles** of trimethylamine.

STEP 4

Similarly, we **convert** the volume of HClO4\text{HClO}_4 to liters (39.1 mL=0.0391 L39.1 \text{ mL} = 0.0391 \text{ L}) and **multiply** by its molarity: 0.0391 L⋅0.7800molL=0.0305 mol0.0391 \text{ L} \cdot 0.7800 \frac{\text{mol}}{\text{L}} = \mathbf{0.0305} \text{ mol} We've added **0.0305 moles** of HClO4\text{HClO}_4.

STEP 5

The added acid reacts with the base.
Since the reaction is 1:1, we **subtract** the moles of added acid from the initial moles of base: 0.111144 mol−0.0305 mol=0.080644 mol0.111144 \text{ mol} - 0.0305 \text{ mol} = \mathbf{0.080644} \text{ mol} We now have **0.080644 moles** of trimethylamine left.

STEP 6

The reaction between trimethylamine and HClO4\text{HClO}_4 forms the conjugate acid, (CH3)3NH+(CH_3)_3NH^+.
Since the reaction is 1:1, the moles of conjugate acid formed **equals** the moles of HClO4\text{HClO}_4 added: **0.0305 moles**.

STEP 7

We're given the pKbpK_b of trimethylamine as 4.19.

STEP 8

We can now use the Henderson-Hasselbalch equation to find the pOH: pOH=pKb+log⁡[conjugate acid][base]pOH = pK_b + \log\frac{[\text{conjugate acid}]}{[\text{base}]} Since the final volume is the same for both the base and the conjugate acid, we can use moles instead of concentrations: pOH=4.19+log⁡0.03050.080644pOH = 4.19 + \log\frac{0.0305}{0.080644} pOH=4.19+log⁡(0.3782)pOH = 4.19 + \log(0.3782)pOH=4.19−0.4225pOH = 4.19 - 0.4225pOH=3.7675pOH = \mathbf{3.7675}The pOH is **3.77**.

STEP 9

Finally, we can **calculate** the pH using the relationship between pH and pOH: pH+pOH=14pH + pOH = 14 pH=14−pOHpH = 14 - pOHpH=14−3.7675pH = 14 - 3.7675pH=10.2325pH = \mathbf{10.2325}

STEP 10

The pH of the solution after adding the HClO4\text{HClO}_4 is **10.23**.

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