What is the pH of a solution formed by combining 50.00 mL of a 0.0855 M acetic a

Question

What is the pH of a solution formed by combining 50.00 mL of a 0.0855 M acetic acid solution, 25.00 mL of 0.1061 M NaOH solution and 25.00 mL of deionized water? Show your work. What is the pH when the solution in 1a is diluted 1 mL in 10 mL total volume? Show your work. What is the pH when 25.00 mL of the solution in #1a has 5.00 mL of a 0.1056 M HCI solution added to it? Show your work. What is the pH when 25.00 mL of the solution in #1a has 5.00 mL of a 0.1056 M NaOH solution added to it? Show your work. What volume of a 0.1048 M NaOH is needed to neutralize (get to the equivalence point) 50.00 mL of a 0.0876 M acetic acid solution? What is the pH at the equivalence point? Show your work. In the titration of 15.00 mL of a 0.1000 M acetic acid solution by a 0.1000 M NaOH solution, find the pH of the solution being titrated when a) 0.00 mL of NaOH has been added, b) 5.00 mL of NaOH has been added, c) 7.50 mL of NaOH has been added d) 14.90 mL of NaOH has been added e) 15.00 mL of NaOH has been added f) 30.00 mL of NaOH has been added Show your work!

Explanation / Answer

a)

Calculate mol of acetic acid -

50.00 mL of 0.0855 M acetic acid

Mol of acetic acid = 0.0855 x 50/1000 = 0.004275 mol

Calculate mol of NaOH-

25.00 mL of 0.1061 M NaOH

0.002652 mol of NaOH = 0.1061 x 25/1000 = 0.002652mol

The reaction between acetic acid and NaOH is-

CH3COOH + NaOH ==> CH3COONa + H2O

This reaction equation indicates that one mole of acetic acid reacts with one mole of NaOH.

Hence 0.002652 mol of NaOH reacts with 0.002652 mol of CH3COOH

So, 0.004275 - 0.002652 = 0.001623 mol 0.00100 mol of CH3COOH left in a total of 100.00 ml

Molarity of CH3COOH = 0.001623 mol/100ml /1000 = 0.01623M

Molarity of CH3COONa = 0.002652/100/1000 = 0.02652M

Using the Henderson-Hasselbalch to calculate the pH -

pH = pKa + log ( [base]/[acid])

Ka for acetic acid is 1.8x10^-5 thus pKa is = 4.74

pH= 4.74 + log (0.02652/0.01623)

pH= 4.74 + 0.21

pH= 4.95

b)

Moles solute before dilution = moles solute after dilution

For CH3COOH

M1V1 = M2V2

1ml x 0.01623M = M2 x 9 ml

M2 = 0.01623M/9 = 0.0018M

For CH3COONa

M2 = 0.02652M/9 = 0.00295

Using the Henderson-Hasselbalch to calculate the pH -

pH = pKa + log ( [base]/[acid])

Ka for acetic acid is 1.8x10^-5 thus pKa is = 4.74

pH= 4.74 + log (0.00295/0.0018)

pH= 4.74 + 0.215

pH = 4.955

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