Question

question 42 (2 points)
instructions: answer all of the following questions on a separate piece of paper. then scan and attach the document by clicking the insert stuff icon. please put 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)
2. 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)
3. 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:

Sub - question 1

Step 1: Define the ionization of methanoic acid

Methanoic acid (\(HCOOH\)) is a weak acid, and its ionization in water is given by \(HCOOH(aq)+H_2O(l)
ightleftharpoons HCOO^-(aq)+H_3O^+(aq)\). Let the concentration of \(H_3O^+\) ions formed be \(x\) mol/L. At equilibrium, the concentration of \(HCOOH\) will be \((0.35 - x)\) mol/L, and the concentrations of \(HCOO^-\) and \(H_3O^+\) will be \(x\) mol/L each. The ionization constant \(K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}\). Since \(K_a = 1.8\times10^{-4}\) and for weak acids, if \(K_a\times100<\frac{C}{K_a}\) (where \(C\) is the initial concentration of the acid), we can approximate \(0.35 - x\approx0.35\). So, \(K_a=\frac{x\cdot x}{0.35}\).

Step 2: Solve for \(x\)

We have the equation \(1.8\times10^{-4}=\frac{x^{2}}{0.35}\). Cross - multiplying gives \(x^{2}=1.8\times10^{-4}\times0.35 = 6.3\times10^{-5}\). Taking the square root of both sides, \(x=\sqrt{6.3\times10^{-5}}\approx7.94\times10^{-3}\) mol/L.

Step 3: Calculate the pH

The pH is defined as \(pH =-\log[H_3O^+]\). Substituting \(x = [H_3O^+]=7.94\times10^{-3}\) into the formula, we get \(pH=-\log(7.94\times10^{-3})\approx2.10\).

Step 1: Write the balanced chemical equation

The reaction between magnesium hydroxide (\(Mg(OH)_2\)) and hydrochloric acid (\(HCl\)) is \(Mg(OH)_2(s)+2HCl(aq)=MgCl_2(aq)+2H_2O(l)\).

Step 2: Calculate the moles of \(Mg(OH)_2\)

The molar mass of \(Mg(OH)_2\) is \(M = 24.31+(16 + 1.01)\times2=58.33\) g/mol. Given the mass of \(Mg(OH)_2\) is \(m = 4.50\) g. The number of moles of \(Mg(OH)_2\), \(n_{Mg(OH)_2}=\frac{m}{M}=\frac{4.50\ g}{58.33\ g/mol}\approx0.0771\) mol.

Step 3: Determine the moles of \(HCl\)

From the balanced equation, 1 mole of \(Mg(OH)_2\) reacts with 2 moles of \(HCl\). So, the moles of \(HCl\), \(n_{HCl}=2\times n_{Mg(OH)_2}=2\times0.0771 = 0.1542\) mol.

Step 4: Calculate the concentration of \(HCl\)

The volume of \(HCl\) solution is \(V = 400.0\ mL=0.400\ L\). The concentration of \(HCl\), \(c=\frac{n_{HCl}}{V}=\frac{0.1542\ mol}{0.400\ L}\approx0.386\) mol/L.

Step 1: Determine the moles of lactic acid and sodium hydroxide at equivalence point

Lactic acid (\(C_3H_6O_3\)) is monoprotic, and the reaction with \(NaOH\) is \(C_3H_6O_3(aq)+NaOH(aq)=C_3H_5O_3Na(aq)+H_2O(l)\). The moles of \(NaOH\) used: \(n_{NaOH}=c\times V=4.0\times10^{-4}\ mol/L\times0.020\ L = 8.0\times10^{-6}\) mol. At the equivalence point, moles of lactic acid (\(n_{acid}\)) = moles of \(NaOH\) (\(n_{base}\)) = \(8.0\times10^{-6}\) mol. The volume of the resulting solution is \(V = 35.0\ mL+20.0\ mL = 55.0\ mL = 0.055\ L\). The concentration of the lactate ion (\(C_3H_5O_3^-\)) formed, \(c=\frac{n}{V}=\frac{8.0\times10^{-6}\ mol}{0.055\ L}\approx1.45\times10^{-4}\) mol/L.

Step 2: Write the hydrolysis reaction of lactate ion

The lactate ion (\(C_3H_5O_3^-\)) hydrolyzes in water: \(C_3H_5O_3^-(aq)+H_2O(l)
ightleftharpoons C_3H_6O_3(aq)+OH^-(aq)\). Let the concentration of \(OH^-\) ions formed be \(x\) mol/L. At equilibrium, \([C_3H_5O_3^-]=(1.45\times10^{-4}-x)\) mol/L, \([C_3H_6O_3]=x\) mol/L, and \([OH^-]=x\) mol/L. The base - ionization constant \(K_b=\frac{K_w}{K_a}\), where \(K_w = 1.0\times10^{-14}\) and \(K_a = 1.4\times10^{-4}\). So, \(K_b=\frac{1.0\times10^{-14}}{1.4\times10^{-4}}\approx7.14\times10^{-11}\).

Step 3: Solve for \(x\)

Since \(K_b\) is very small, we can approximate \(1.45\times10^{-4}-x\approx1.45\times10^{-4}\). Then \(K_b=\frac{[C_3H_6O_3][OH^-]}{[C_3H_5O_3^-]}=\frac{x\cdot x}{1.45\times10^{-4}}\). So, \(x^{2}=K_b\times1.45\times10^{-4}=7.14\times10^{-11}\times1.45\times10^{-4}\approx1.035\times10^{-14}\). Taking the square root, \(x=\sqrt{1.035\times10^{-14}}\approx1.02\times10^{-7}\) mol/L.

Step 4: Calculate the pOH and then pH

\(pOH=-\log[OH^-]=-\log(1.02\times10^{-7})\approx6.99\). Since \(pH + pOH=14\), \(pH = 14 - 6.99 = 7.01\).

Answer:

\(2.10\)

Sub - question 2