Question

solve the following ph problems

a) if the ph is 10.85 and you have 3.05 l of solution, how many grams of magnesium hydroxide, mg(oh)₂, are in the solution? assume full dissociation. assume this involves a strong base. the equation is: mg(oh)₂ → mg⁺² + 2oh⁻¹

b) at a particular temperature, a 0.195 m hcl solution has a ph = 1.85. calculate the value of kₐ at this temperature. the equation is: hcl + h₂o ⇌ cl⁻¹ + h₃o⁺¹

c) calculate the h₃o⁺¹ in 0.55 m no⁻³ (kₐ = 3.5 x 10⁻¹⁰) the equation is: no⁻³ + h₂o ⇌ hno₃ + oh⁻¹

d) the ph of a 0.950 m solution of the weak base z⁻¹ is 12.35. what is the k_b for z⁻¹? the equation is: z⁻¹ + h₂o ⇌ hz + oh⁻¹

Explanation:

Part (a)

Step 1: Find pOH from pH

We know that \( \text{pH} + \text{pOH} = 14 \). Given \( \text{pH} = 10.85 \), so \( \text{pOH} = 14 - 10.85 = 3.15 \)

Step 2: Find \([\text{OH}^-]\) from pOH

Using the formula \( \text{pOH} = -\log[\text{OH}^-] \), we can find \([\text{OH}^-]\) as \( [\text{OH}^-] = 10^{-\text{pOH}} \). Substituting \( \text{pOH} = 3.15 \), we get \( [\text{OH}^-] = 10^{-3.15} \approx 7.08 \times 10^{-4} \, \text{M} \)

Step 3: Relate \([\text{OH}^-]\) to \([\text{Mg(OH)}_2]\)

From the dissociation equation \( \text{Mg(OH)}_2
ightarrow \text{Mg}^{2+} + 2\text{OH}^- \), we see that 1 mole of \( \text{Mg(OH)}_2 \) produces 2 moles of \( \text{OH}^- \). So, \( [\text{Mg(OH)}_2] = \frac{[\text{OH}^-]}{2} \). Substituting \( [\text{OH}^-] = 7.08 \times 10^{-4} \, \text{M} \), we get \( [\text{Mg(OH)}_2] = \frac{7.08 \times 10^{-4}}{2} \approx 3.54 \times 10^{-4} \, \text{M} \)

Step 4: Calculate moles of \( \text{Mg(OH)}_2 \)

Moles \( = \text{Molarity} \times \text{Volume} \). Volume \( = 3.05 \, \text{L} \), so moles of \( \text{Mg(OH)}_2 = 3.54 \times 10^{-4} \, \text{M} \times 3.05 \, \text{L} \approx 1.08 \times 10^{-3} \, \text{mol} \)

Step 5: Calculate mass of \( \text{Mg(OH)}_2 \)

Molar mass of \( \text{Mg(OH)}_2 \) is \( 24.31 + 2(16.00 + 1.01) = 58.33 \, \text{g/mol} \). Mass \( = \text{Moles} \times \text{Molar Mass} \). So, mass \( = 1.08 \times 10^{-3} \, \text{mol} \times 58.33 \, \text{g/mol} \approx 0.0630 \, \text{g} \)

Step 1: Find \([\text{H}_3\text{O}^+]\) from pH

Given \( \text{pH} = 1.85 \), using \( \text{pH} = -\log[\text{H}_3\text{O}^+] \), we get \( [\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-1.85} \approx 0.0141 \, \text{M} \)

Step 2: Analyze the dissociation of HCl

The equation is \( \text{HCl} + \text{H}_2\text{O}
ightleftharpoons \text{Cl}^- + \text{H}_3\text{O}^+ \). Let the initial concentration of HCl be \( c = 0.195 \, \text{M} \). At equilibrium, \( [\text{H}_3\text{O}^+] = [\text{Cl}^-] = x = 0.0141 \, \text{M} \) (from pH), and \( [\text{HCl}] = c - x = 0.195 - 0.0141 = 0.1809 \, \text{M} \)

Step 3: Calculate \( K_a \)

The expression for \( K_a \) is \( K_a = \frac{[\text{Cl}^-][\text{H}_3\text{O}^+]}{[\text{HCl}]} \). Substituting the values, we get \( K_a = \frac{(0.0141)(0.0141)}{0.1809} \approx \frac{0.00019881}{0.1809} \approx 1.10 \times 10^{-3} \)

Step 1: Recognize the reaction type

The ion \( \text{NO}_3^- \) is the conjugate base of \( \text{HNO}_3 \) (a strong acid), so \( \text{NO}_3^- \) does not hydrolyze significantly. But wait, the given \( K_a = 3.5 \times 10^{-10} \) is for \( \text{HNO}_3 \)? Wait, no, maybe it's a typo and it's \( \text{NO}_2^- \) (nitrite ion) which is the conjugate base of \( \text{HNO}_2 \) (weak acid). Assuming it's \( \text{NO}_2^- \), let's proceed. The reaction is \( \text{NO}_2^- + \text{H}_2\text{O}
ightleftharpoons \text{HNO}_2 + \text{OH}^- \). First, find \( K_b \) from \( K_a \) of \( \text{HNO}_2 \). We know that \( K_w = K_a \times K_b \), so \( K_b = \frac{K_w}{K_a} \). \( K_w = 1.0 \times 10^{-14} \), \( K_a = 3.5 \times 10^{-10} \), so \( K_b = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-10}} \approx 2.86 \times 10^{-5} \)

Step 2: Set up the ICE table

Let \( x = [\text{OH}^-] = [\text{HNO}_2] \) at equilibrium. Initial concentration of \( \text{NO}_2^- \) is \( 0.55 \, \text{M} \). So,

Species	Initial (M)	Change (M)	Equilibrium (M)
\( \text{HNO}_2 \)	0	+x	x
\( \text{OH}^- \)	0	+x	x

Step 3: Write the \( K_b \) expression

\( K_b = \frac{[\text{HNO}_2][\text{OH}^-]}{[\text{NO}_2^-]} = \frac{x \cdot x}{0.55 - x} \). Since \( K_b \) is small, \( 0.55 - x \approx 0.55 \). So, \( 2.86 \times 10^{-5} = \frac{x^2}{0.55} \)

Step 4: Solve for x ( \( [\text{OH}^-] \) )

\( x^2 = 2.86 \times 10^{-5} \times 0.55 \approx 1.573 \times 10^{-5} \), so \( x = \sqrt{1.573 \times 10^{-5}} \approx 3.97 \times 10^{-3} \, \text{M} \) ( \( [\text{OH}^-] \) )

Step 5: Find \( [\text{H}_3\text{O}^+] \)

Using \( K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \), we get \( [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{3.97 \times 10^{-3}} \approx 2.52 \times 10^{-12} \, \text{M} \)

Answer:

\( \approx 0.0630 \, \text{g} \)

Part (b)