Problem: You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA?

FREE Expert Solution

HA (aq) + H2O ⇌ H3O+ (aq) + A-(aq)

(1) Calculate the moles of Hand OH-

mol HA = 75 mL (1x10-3 L1 mL)(0.10 mol HA1 L)

mol HA = 0.0075 mol HA

mol OH- = 30 mL (1x10-3 L1 mL)(0.10 mol NaOH 1 L)(1 mol OH-1 mol NaOH)

mol OH- = 0.003 mol OH-


(2) Calculate excess HA and its molarity

0.0075 - 0.003 = 0.0045 mol

[HA] = 0.0045 mol(75 +30) mL (1x10-3 L1 mL)

[HA] = 0.0429 M

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Problem Details

You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA?

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Our tutors have indicated that to solve this problem you will need to apply the Acid and Base Titration Curves concept. You can view video lessons to learn Acid and Base Titration Curves. Or if you need more Acid and Base Titration Curves practice, you can also practice Acid and Base Titration Curves practice problems.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.