An aqueous solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂)₂] = 0.200 M.K_a = 6.3 × 10^-5 for HC₇H₅O₂.The solution volume is 5.00 L.What is the pH of the solution after 10.00 mL of 5.00 M NaOH is added? A)4.80 B)4.86 C)4.65 D)4.70 E)4.75

Question

Quizplus · Accepted Answer

First, let's write the equation for the dissociation of HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0:

HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0(aq) + H2O(l) ⇌ H3O+(aq) + CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0-(aq)

The K for this reaction is given as 6.3 × 10S1U1P1-5S1S1P0.

Next, we need to determine if adding NaOH will result in a significant amount of the HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 dissociating. We can do this by calculating the initial concentration of H3O+:

[H3O+] = K × [HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0] = (6.3 × 10S1U1P1-5S1S1P0)(0.100) = 6.3 × 10S1U1P6S1S1P0

This is a relatively low concentration of H3O+, so we can safely assume that adding NaOH will result in only a small amount of the HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 dissociating.

Now, we can write the equation for the reaction between NaOH and H3O+:

NaOH(aq) + H3O+(aq) → H2O(l) + Na+(aq)

We can use the equation for the reaction of a strong acid and strong base to determine the pH after 10.00 mL of 5.00 M NaOH is added:

nbase × Vbase = nacid × Vacid

Here, nbase is the number of moles of NaOH added, Vbase is the volume of NaOH added, nacid is the number of moles of H3O+ in the solution prior to the addition of NaOH (which we just calculated to be 6.3 × 10S1U1P6S1S1P0), and Vacid is the volume of the entire solution (5.00 L). Solving for nbase, we get:

nbase = nacid × Vacid / Vbase = (6.3 × 10S1U1P6S1S1P0)(5.00 L) / (10.00 × 10S1U1P3S1S1P0 L) = 3.15 × 10S1U1P0S1S1P0 mol

This amount of NaOH will fully react with all of the H3O+ in the solution, resulting in the formation of 3.15 × 10S1U1P0S1S1P0 mol of water.

Now we can calculate the new concentration of the anion:

[CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0-] = [Ca(CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0)S1U1B12S1U1B0] / (Vacid + Vbase)

= (0.200 M) / (5.00 L + 10.00 × 10S1U1P3S1S1P0 L)

= 1.33 × 10S1U1P2S1S1P0 M

We can use this concentration to calculate the new concentration of the acid:

[HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0] = K × [CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0-] / [H3O+]

= (6.3 × 10S1U1P1-5S1S1P0)(1.33 × 10S1U1P2S1S1P0) / (1.87 × 10S1U1P6S1S1P0)

= 4.46 × 10S1U1P-4S1S1P0 M

Finally, we can calculate the pH using the equation for weak acid dissociation:

pH = pKa + log([A-] / [HA])

= 3.20 + log(1.33 × 10S1U1P2S1S1P0 / 4.46 × 10S1U1P-4S1S1P0)

= 4.86

Therefore, the pH after 10.00 mL of 5.00 M NaOH is added is 4.86 (choice B).

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