Please help i have no clue what to do For the next three problems, consider 1.0

Question

Please help i have no clue what to do

For the next three problems, consider 1.0 L of a solution which is 0.6 M HC2H3O2 and 0.2 M NaC2H3O2 (Ka for HC2H3O2 = 1.8 x 10-5). Assume 2 significant figures in all of the given concentrations so that you should calculate all of the following pH values to two decimal places.

1. Calculate the pH of this solution.

2. Calculate the pH after 0.10 mol of HCl has been added to the original solution. Assume no volume change on addition of HCl.

3. Calculate the pH after 0.20 mol of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

Explanation / Answer

1) the given solution is a buffer solution

we know that

for buffers

pH = pKa + log [salt / acid ]

so

pH = -log Ka + log [NaC2H302 / HC2H302]

so

using given values

pH = -log 1.8 x 10-5 + log [0.2/0.6]

pH = 4.27

so

the pH is 4.27

2)

now

we know that

moles = molarity x volume

given volume is 1 L

so

moles of HC2H3O2 = 0.6 x 1 = 0.6

moles of NaC2H3O2 = 0.2 x 1 = 0.2

now

0.1 mol of HCL is added

the reaction becomes

NaC2H3O2 + HCL ---> HC2H3O2 + NaCl

moles of NaC2H3O2 reacted = moles of hCL added = 0.1

moles of NaC2H3O2 remaining = 0.2 - 0.1 = 0.1

now

moles of HC2H3O2 formed = moles of HCl added = 0.1

new moles of HC2H3O2 = 0.1 + 0.6 = 0.7

now

pH = -log Ka + log [NaC2H302 / HC2H302]

so

using given values

pH = -log 1.8 x 10-5 + log [0.1/0.7]

pH = 3.90

so the pH becomes 3.90

3)

now

0.20 mol of NaOH is added

the reaction becomes

HC2H3O2 + NaOH ---> NaC2H3O2 + H20

moles of HC2H3O2 reacted = moles of NaOH added = 0.2

moles of HC2H3O2 remaining = 0.6 - 0.2 = 0.4

now

moles of NaC2H3O2 formed = moles of NaOH added = 0.2

new moles of NaC2H3O2 = 0.2 + 0.2 = 0.4

now

pH = -log Ka + log [NaC2H302 / HC2H302]

so

using given values

pH = -log 1.8 x 10-5 + log [0.4/0.4]

pH = 4.74

so the pH becomes 4.74

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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