Part 3, Solutions B and C Prepare 20.00 ml of the two utions as described below,

Question

Part 3, Solutions B and C Prepare 20.00 ml of the two utions as described below, following the procedure as for Solution Predict the pH then measure the pH Solution B: 10.00 mLof0. acid and M sodium hydroxide Solution C: 10.00 ml of 02 M sodium acetate and 10.00 mL of 0.1 M Hc1 To solution B, add .2 mL ero.1 M Hchas was done for Soluti A. Measure and record the pH To solution c, add 1 2 mt. of 0.1 M NaoHL Measure the pH and record. Address the following in the hitial Results and Conclusions: solution1o the addition of acid or base, are these sol Based on the response of each buffers? 3.2 write a halanced equation for the reaction showing the formation of each solution. 3.3 Without doing extensive calculations, are the compositions of solutions B and c the 3.4 same when initially mixed constitute the buffer. buffer is formed, identify the components that the change in pH consistenl Explain the added is handled by this explanationT 3.6 buse is handled by solution c. ls the change in pH comsistent with this explanation? 3.7 Describe three ways to make a buffer, for example, adding excess strong acid to a strong base (this is NOT a way to make a buffer).

Explanation / Answer

Solution B : 10 ml of 0.1 M AcOH + 10 ml of 0.1 M NaOH

we would have all of acid neutralized by added base,

[NaOAc] formed = 0.1 M x 10 ml/20 ml = 0.05 M

So we have 0.05 M of 20 ml NaOAc as solution B

For initial pH,

1 x 10^-14/1.8 x 10^-5 = x^2/0.05

x = [OH-] = 5.27 x 10^-6 M

pOH = -log[OH-] = 5.28

pH = 14 - pOH = 8.72

When, 1-2 ml of 0.1 M HCl is added,

we would have a buffer solution formed of AcOH (formed due to HCl addition) and remaining NaOAc

Here NaOAc would be in excess

Solution C : 10 ml of 0.1 M NaOAc + 10 ml of 0.1 M HCl

we would have all of salt NaOAc neutralized by added acid,

[AcOH] formed = 0.1 M x 10 ml/20 ml = 0.05 M

So we have 0.05 M of 20 ml AcOH as solution C

For initial pH,

1.8 x 10^-5 = x^2/0.05

x = [H3O+] = 9.50 x 10^-4 M

pH = -log[H3O+] = 3.023

When, 1-2 ml of 0.1 M NaOH is added,

we would have a buffer solution formed of NaOAc (formed due to NaOH addition) and remaining AcOH

Here AcOH would be in excess

3.1 Both solution B and solution C would form buffer when mixed with acid and base.

3.2 Solution B,

NaOAc + HCl <==> AcOH + NaCl

Solution C,

AcOH + NaOH <==> NaOAc + H2O

3.3 No the composition of solution B and C are different, solution B is for NaOAc (0.05 M) and solution C is for AcOH (0.05 M)

3.4 Buffer formed when solution B + HCl = NaOAc/AcOH

Buffer formed when solution C + NaOH = AcOH/NaOAc

3.5 The added acid reacts with NaOAc in solution B to form AcOH to form buffer solution

the pH would be lower than initial pH

3.6 The added acid reacts with AcOH and forms NaOAc to form buffer solution

the pH would be higher than initial pH

3.7 Methods to generate buffers

Mix weak acid with strong base

or,

Mix weak base with strong acid

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