Question

Neue Suit II. Preparation of Buffers and Buffer Capacity A. Preparation of Buffer Solutions • Buffer Solution 1: Volume, mL Volume, L HC4H10 | 40,0 lo.cu. NaOH 20.0 0.02 OH + HC,H, O H Molarity 0.100 0.100 ,00 + (V x mol/L) Moles 0.004 0.002 CHO lect - mol mol - mol mol + mol mol mol mol mol (See Study Guide Key Problem 3 ) Total Volume of Buffer Solution - 30.0 mL-0.08 L. Molarity of HC,H,O2 = HA- _M (mols HC H302 / total vol (L) Molarity of CHO = A= M (mols C2H2O2 / total vol (L) pH = 4,74 + 109 → PH - 4.74 + log pH- Note: Calculations for Buffer Solution 2 similar to Buffer Solution 1 but with different volumes. • Buffer Solution 2: Volume, mL Volume, Molarity Moles HCH 01 | 40,0 0 04 | 0.100 0.004 NaOH 30.0 0.03 0.100 0.003 B | 0H (al + HCHC (20 = 10 ml + C3H03 |8g| m ol mol 0 mol mol - mol - + mol mol mol - mol Total Volume of Buffer Solution 80.0_mL - 0.08 Molarity of HCH.0, = HA = Molarity of CH.0; = = pH = 4.74 + loga pH = 4.74 +log pH- B. pH Measurement of B Type here to Et os

need the ice table filled out along with the questions below. This is for an lab that was converted to online, and I'm not sure how to complete the assignment

Answer 1

Balanced Chemical Equation:-

OH-(aq) + HC2H3O2(aq) -----> H2O(l) + C2H3O2-(aq)

It is evident from the balanced chemical equation that 1 mole of OH- ion reacts with 1 mole of HC2H3O2 to produce 1 mole of H2O and 1 mole of C2H3O2- ion.

BUFFER SOLUTION 1:

Initial moles of HC2H3O2= 0.004 moles

Initial moles of NaOH or OH- ion= 0.002 moles

Here NaOH will act as a limiting reagent because it is in a lesser amount than HC2H3O2. Hence, 0.002 moles of OH- will react with 0.002 moles of HC2H3O2, leaving behind 0.002 moles of unreacted HC2H3O2 and producing 0.002 moles of C2H3O2-. Depending upon this explanation ICE Table is formed below.

	OH-(moles)	HC2H3O2(moles)	H2O(moles)	C2H3O2-(moles)
I	0.002	0.004	-	0
C	-0.002	-0.002	-	+0.002
E	0	0.002	-	0.002

Here,

I represent the initial concentration.

C represents the change in concentration.

E represents the final/equilibrium concentration.

Total volume of Buffer Solution= 0.08 L

Molarity of HC2H3O2= HA = (moles of HC2H3O2)/(total volume)

=0.002/0.08 = 0.025

HA = 0.025 M

Molarity of C2H3O2-= A- = (moles of C2H3O2-​​​​​​​)/(total volume)

=0.002/0.08 = 0.025

A- = 0.025 M

pH= 4.74 + log(A-/HA)

= 4.74 + log(0.025/0.025)

= 4.74 + log(1)

= 4.74 + 0

= 4.74

pH= 4.74

BUFFER SOLUTION 2:(following the same procedure as above)

Initial moles of HC2H3O2= 0.004 moles

Initial moles of NaOH or OH- ion= 0.003 moles

Here NaOH will act as a limiting reagent because it is in a lesser amount than HC2H3O2. Hence, 0.003 moles of OH- will react with 0.003 moles of HC2H3O2, leaving behind 0.001 moles of unreacted HC2H3O2 and producing 0.003 moles of C2H3O2-. Depending upon this explanation ICE Table is formed below.

	OH-(moles)	HC2H3O2(moles)	H2O(moles)	C2H3O2-(moles)
I	0.003	0.004	-	0
C	-0.003	-0.003	-	+0.003
E	0	0.001	-	0.003

Here,

I represent the initial concentration.

C represents the change in concentration.

E represents the final/equilibrium concentration.

Total volume of Buffer Solution= 0.08 L

Molarity of HC2H3O2= HA = (moles of HC2H3O2​​​​​​​)/(total volume)

=0.001/0.08 = 0.0125

HA = 0.0125 M

Molarity of C2H3O2-= A- = (moles of C2H3O2-​​​​​​​)/(total volume)

=0.003/0.08 = 0.0375

A- = 0.0375 M

pH= 4.74 + log(A-/HA)

= 4.74 + log(0.0125/0.0375)

= 4.74 + log(1/3)

= 4.74 - 0.4711

= 4.2629

pH= 4.26