Question

question 42 (2 points)
instructions: answer all of the following questions on a separate piece of paper.
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your name on your scanned image.

1. methanoic acid is also called formic acid. it has the

chemical formula hcooh(l). it is a colourless fuming liquid
that is mainly used as a preservative. it exhibits the
following equilibrium in water:
hcooh(aq) + h₂o(l) ⇌ hcoo⁻(aq) + h₃o⁺(aq)

if the ionization constant for the above acid is 1.8 × 10⁻⁴, the ph of methanoic acid
in a 0.35 mol/l solution will be ______.
(2 marks)

1. a 4.50 g tablet of magnesium hydroxide neutralizes 400.0 ml of stomach acid,

hcl. what is the concentration of hcl in the stomach? (2 marks)

1. a 35.0 ml sample of (monoprotic) lactic acid, c₃h₆o₃, is titrated with 20.0 ml of a

4.0 × 10⁻⁴ mol/l sodium hydroxide solution. what is the ph of the resulting solution
at the equivalence point, if kₐ for lactic acid is 1.4 × 10⁻⁴? (9 marks)

1. the concentration of the hydroxide ion, oh⁻(aq), in an aqueous solution is

2.5 × 10⁻³ mol/l. what is the ph of the solution? (2 marks)

Explanation:

1) Solution for Methanoic Acid pH Calculation

Step1: Define the ionization of HCOOH

Methanoic acid (HCOOH) ionizes as: $\ce{HCOOH(aq) + H_{2}O(l) <=> HCOO^{-}(aq) + H_{3}O^{+}(aq)}$. Let the concentration of $\ce{H_{3}O^{+}}$ at equilibrium be $x$. Then, $[\ce{HCOO^{-}}] = x$ and $[\ce{HCOOH}] = 0.35 - x$. Since $K_a$ is small ($1.8\times10^{-4}$), we can approximate $0.35 - x \approx 0.35$.

Step2: Use the $K_a$ expression

The ionization constant $K_a = \frac{[\ce{HCOO^{-}}][\ce{H_{3}O^{+}}]}{[\ce{HCOOH}]}$. Substituting the values: $1.8\times10^{-4} = \frac{x\cdot x}{0.35}$.

Step3: Solve for $x$

$x^{2} = 1.8\times10^{-4} \times 0.35 = 6.3\times10^{-5}$. So, $x = \sqrt{6.3\times10^{-5}} \approx 7.94\times10^{-3}\ \text{mol/L}$. This $x$ is $[\ce{H_{3}O^{+}}]$.

Step4: Calculate pH

$\text{pH} = -\log([\ce{H_{3}O^{+}}]) = -\log(7.94\times10^{-3}) \approx 2.10$.

Step1: Write the balanced reaction

The reaction between $\ce{Mg(OH)_{2}}$ and $\ce{HCl}$ is: $\ce{Mg(OH)_{2}(s) + 2HCl(aq) -> MgCl_{2}(aq) + 2H_{2}O(l)}$. Molar mass of $\ce{Mg(OH)_{2}}$ is $24.31 + 2\times(16.00 + 1.01) = 58.33\ \text{g/mol}$.

Step2: Calculate moles of $\ce{Mg(OH)_{2}}$

Moles of $\ce{Mg(OH)_{2}} = \frac{4.50\ \text{g}}{58.33\ \text{g/mol}} \approx 0.0771\ \text{mol}$.

Step3: Determine moles of HCl

From the reaction, 1 mol of $\ce{Mg(OH)_{2}}$ reacts with 2 mol of HCl. So, moles of HCl $= 2\times0.0771 = 0.1542\ \text{mol}$.

Step4: Calculate HCl concentration

Volume of HCl is $400.0\ \text{mL} = 0.4000\ \text{L}$. Concentration of HCl $= \frac{0.1542\ \text{mol}}{0.4000\ \text{L}} \approx 0.3855\ \text{mol/L}$.

Step1: Determine moles of lactic acid and NaOH at equivalence

Moles of $\ce{NaOH} = C\times V = 4.0\times10^{-4}\ \text{mol/L} \times 0.0200\ \text{L} = 8.0\times10^{-6}\ \text{mol}$. At equivalence, moles of lactic acid ($\ce{HLac}$) = moles of $\ce{NaOH} = 8.0\times10^{-6}\ \text{mol}$. Volume of solution $= 35.0\ \text{mL} + 20.0\ \text{mL} = 55.0\ \text{mL} = 0.0550\ \text{L}$.

Step2: Concentration of lactate ion ($\ce{Lac^{-}}$)

After titration, $\ce{HLac}$ is converted to $\ce{Lac^{-}}$. Concentration of $\ce{Lac^{-}} = \frac{8.0\times10^{-6}\ \text{mol}}{0.0550\ \text{L}} \approx 1.45\times10^{-4}\ \text{mol/L}$.

Step3: Hydrolysis of $\ce{Lac^{-}}$

$\ce{Lac^{-}(aq) + H_{2}O(l) <=> HLac(aq) + OH^{-}(aq)}$. Let $[\ce{OH^{-}}] = x$, then $[\ce{HLac}] = x$ and $[\ce{Lac^{-}}] = 1.45\times10^{-4} - x$. $K_b = \frac{K_w}{K_a} = \frac{1.0\times10^{-14}}{1.4\times10^{-4}} \approx 7.14\times10^{-11}$.

Step4: Use $K_b$ expression

$K_b = \frac{[\ce{HLac}][\ce{OH^{-}}]}{[\ce{Lac^{-}}]} \approx \frac{x\cdot x}{1.45\times10^{-4}}$. So, $x^{2} = 7.14\times10^{-11} \times 1.45\times10^{-4} \approx 1.035\times10^{-14}$. Then, $x = \sqrt{1.035\times10^{-14}} \approx 1.017\times10^{-7}\ \text{mol/L}$. This is $[\ce{OH^{-}}]$.

Step5: Calculate pOH and pH

$\text{pOH} = -\log(1.017\times10^{-7}) \approx 6.99$. $\text{pH} = 14 - 6.99 = 7.01$. (Note: Due to approximations, the value might vary slightly, but around 7.0 - 7.02 is acceptable.)

Answer:

The pH of the methanoic acid solution is approximately $\boldsymbol{2.10}$.

2) Solution for HCl Concentration (Magnesium Hydroxide Neutralization)