Question

what is the ph of a solution that results from mixing 25.0 ml of0.200 m ha with 12.5 ml of 0.400 m naoh? (ka = 1.0x 1 o-5)

Answer 1

The pH of the resulting solution is approximately 4.834.

To find the pH of the solution resulting from mixing 25.0 mL of 0.200 M HA (a weak acid) with 12.5 mL of 0.400 M NaOH (a strong base), you can follow these steps:

Step 1: Calculate the moles of HA and NaOH in the respective solutions.

Moles of HA = (Volume of HA solution in liters) × (Concentration of HA)

Moles of HA = (25.0 mL / 1000 mL/L) × 0.200 M = 0.005 moles

Moles of NaOH = (Volume of NaOH solution in liters) × (Concentration of NaOH)

Moles of NaOH = (12.5 mL / 1000 mL/L) × 0.400 M = 0.005 moles

Step 2: Determine the limiting reactant. Since both HA and NaOH have the same number of moles (0.005 moles), neither is in excess. However, HA is a weak acid, and NaOH is a strong base, so they will react in a 1:1 ratio to form water and the conjugate base A-.

HA + NaOH → H2O + A-

Step 3: Calculate the concentration of A- (the conjugate base) formed in the reaction.

The initial concentration of A- is zero, and we have formed 0.005 moles of A-. We can find the final volume of the solution by adding the volumes of the two solutions:

Total volume = Volume of HA solution + Volume of NaOH solution

Total volume = 25.0 mL + 12.5 mL = 37.5 mL = 0.0375 L

Now, calculate the concentration of A-:

Concentration of A- = (moles of A-) / (total volume)

Concentration of A- = (0.005 moles) / (0.0375 L) = 0.1333 M

Step 4: Use the equilibrium expression and the given Ka value to calculate the concentration of H+ (protons) in the solution:

HA ⇌ H+ + A-

Ka = [H+][A-] / [HA]

We already know the concentration of A- is 0.1333 M, and we can set up the equation with the initial concentration of HA (0.200 M - the concentration of HA that reacted):

1.0 x 10^-5 = [H+][0.1333] / (0.200 - 0.005)

1.0 x 10^-5 = [H+][0.1333] / 0.195

Now, solve for [H+]:

[H+] = (1.0 x 10^-5) × (0.195 / 0.1333)

[H+] ≈ 1.464 x 10^-5 M

Step 5: Calculate the pH using the formula:

pH = -log10[H+]

pH = -log10(1.464 x 10^-5)

pH ≈ 4.834

So, the pH of the resulting solution is approximately 4.834.

Answer 2

The pH of the solution is approximately 13.12 and here is how:

The balanced equation:

`HA + NaOH --- > NaA + H_2 O`

First, let's calculate the moles of each solution:

`\(\text{moles of } HA = (0.200 \, \text{mol/L}) * (0.0250 \, \text{L}) = 0.005 \, \text{mol}\)`

`\(\text{moles of } NaOH = (0.400 \, \text{mol/L}) * (0.0125 \, \text{L}) = 0.005 \, \text{mol}\)`

It is clear that NaOH is the limiting reactant because it reacts in a 1:1 ratio with HA.

Next, let's calculate the moles of the remaining excess reactant (HA):

`\(\text{moles of remaining } HA = \text{moles of } HA - \text{moles of } NaOH = 0.005 \, \text{mol} - 0.005 \, \text{mol} = 0 \, \text{mol}\)`

Then, the concentrations:

`\(\text{concentration of } NaOH = \frac{\text{moles of } NaOH}{\text{total volume in liters}} = \frac{0.005 \, \text{mol}}{0.025 \, \text{L} + 0.0125 \, \text{L}} \approx 0.133 \, \text{mol/L}\)`

`\(\text{concentration of } HA = \frac{\text{moles of remaining } HA}{\text{total volume in liters}} = \frac{0 \, \text{mol}}{0.025 \, \text{L} + 0.0125 \, \text{L}} = 0 \, \text{mol/L}\)`

Write the expression for Ka:

`K_a = ([H_3O^+][A^-])/([HA])`

Also

`\(HA \rightleftharpoons H^+ + A^-\)`

Now, let's substitute the concentrations into the Ka expression:

`\(1.0 * 10^(-5) = \frac{{[H_3O^+][A^-]}}{\text{[HA]}}\)`

Since NaOH is the limiting reactant, [A⁻] is the same as the moles of NaOH, which is used up in the reaction.

Now, let's solve for the concentration of \(H₃O^+\):

`{[H_3O^+]} = K_a * \frac{\text{[HA]}}{\text{[A⁻]}} = (1.0 * 10^(-5)) * \frac{0 \, \text{mol/L}}{0.133 \, \text{mol/L}} = 0 \, \text{mol/L}\)`

Now, calculate the pH:

\(\text{pH} = -\log(\text{[H₃O⁺]}) = -\log(0) \)

Since the concentration of
`\(H_3O^+\)` is zero, the pH is undefined. This indicates that the solution is basic, and the concentration of hydroxide ions should be considered. The pH can be found using the relationship
`\(pH + pOH = 14\)`.

`\(\text{pOH} = -\log(\[OH^-})\)`

`{[OH^-]} = \frac{\text{moles of } NaOH}{\text{total volume in liters}} = \frac{0.005 \, \text{mol}}{0.025 \, \text{L} + 0.0125 \, \text{L}} \approx 0.133 \, \text{mol/L}\)`

`\(\text{pOH} = -\log(0.133) \approx 0.88\)`

Thus:
`\(\text{pH} = 14 - 0.88 \approx 13.12\)`

Therefore, the pH of the solution is approximately 13.12.