3.)A certain weak acid, HA, with a K a value of 5.61×10?6, is titrated with NaOH

Question

3.)A certain weak acid, HA, with a Ka value of 5.61×10?6, is titrated with NaOH.

Part A

A solution is made by titrating 7.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH?

Express the pH numerically to two decimal places.

Part B

More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 35.0 mL ?

Express the pH numerically to two decimal places.

5.) Part A

20 mL of 0.25 M of NH3 is titrated with 0.40 M HCl. Calculate the pH of the solution after 20 mL HCl is added. K b (NH3) = 1.8 × 10?5

A.)12.50

B.)7.00

C.)4.74

D.)1.12

A.)12.50

B.)7.00

C.)4.74

D.)1.12

Explanation / Answer

3)

part A)

    Ka = 5.61×10^-6

     pKa = -log Ka

     pKa = 5.25

HA    +    NaOH ------------------------> NaA + H2O

7             1                                           0          0

-1          -1                                         +1        +1

6            0                                           1          1

pH = pKa + log [NaA / HA]

pH = 5.25 + log (1 / 6 )

pH = 4.47

part B)

total acid millimoles are converted into salt millimoles

salt concentration = millimoles / total volume

                               = 7 / 35

                               = 0.20 M

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/ 2[5.25 + log 0.2]

pH = 9.28

5) part A)

NH3 millimoles = 20 x 0.25 = 5

HCl millimoles = 20 x 0.40 = 8

[H+]left = 8 - 5 / (20+ 20) = 0.075 M

pH = -log [H+]

pH = -log (0.075)

pH = 1.12

answer : D) 1.12

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.