Question

Addition of Strong Acid or Base to Buffered and Unbuffered Solutions pH after addition of 3.0 M HCI Initial pH pH after 5 drops HCI pH after 10 drops HCI 0.84 pH after 15 drops HCI 0.65 pH after 20 drops HCI 0.56 Solution 1 pH after 25 drops HCI 0.45 2.51 1.18 Solution 2 4.15 3.43 1.32 0.90 0.77 0.64 Solution 3 4.54 4.25 3.77 1.62 1.07 0.81 Solution 4 4.89 4.67 4.40 4.02 2.52 1.23 Solution 5 5.33 5.04 4.78 4.46 4.10 3.45 Solution 6 9.27 5.45 5.11 4.87 4.57 4.22 pH after addition of 3.0 M NaOH Initial pH pH after 5 drops NaOH 3.96 pH after 10 drops NaOH 4.37 pH after 15 drops NaOH 4.64 pH after 20 drops NaOH 4.89 pH after 25 drops NaOH 5.26 Solution 1 2.57 4.14 4.53 4.74 5.00 5.35 6.01 Solution 2 4.81 5.53 12.81 5.08 9.21 4.59 Solution 3 12.42 12.93 13.22 5.24 5.69 4.97 Solution 4 13.39 12.72 13.26 6.04 13.07 5.30 Solution 5 13.42 13.47 12.89 13.16 13.27 9.17 Solution 6

(24pts) Concentration and pH of Buffer Solutions Calculate the concentration of HC,H302 and C2H302 in each buffer solution by using the volumes of each solution specified in the experiment protocol. Table view List view Initial Concentrations of Conjugates in Buffer Solutions [HC2H302] (M) [C2H3021 (M) Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6

Table 1 Composition of Solutions in Beakers Brong ghatalystok pucation 2 Beaker # Volume of 0.50 M HC2H:02 Volume of 0.50 M NaC2H302 (mL) (mL) 10.00 0.00 2 8.00 2.00 اليا 4.00 6.00 5 2.00 8.00 6 0.00 10.00

Table view List view Difference between measured pH and calculated theoretical pH of solutions Measured pH Calculated pH ApH (do not include negative sign) Solution 1 2.51 Solution 2 4.15 Solution 3 4.54 Solution 4 4.89 Solution 5 5.33 Solution 6 9.27 inte

1) Find the Difference between measured pH and calculated theoretical pH of solutions 1-6 for [HC₂H₃O₂] (M) and [C₂H₃O₂⁻] (M)

2) Find the Difference between measured pH and calculated theoretical pH of solutions 1-6.

3) In this experiment, we define the buffer capacity of a buffer as the number of drops of either 3.0 M HCl or 3.0 M NaOH needed before the pH of the solution changes by more than 0.5 pH units. For example, if the pH of a buffer solution changes by more than 0.5 pH units during the third 5 drop addition of 3 M HCl then that buffer solution's acid buffer capacity will be "10 drops." Or, if the pH of a buffer solution changes by more than 0.5 pH units during the 1st 5 drop addition of 3 M NaOH then that buffer solution's base buffer capacity is "0 drops. Find for solutions 1-6.

Answer 1

1) (24pts) Concentration and pH of Buffer Solutions Workings: HC2H302 is acetic acid and NaC2H302 is sodium acetate Calculate the concentration of HC H302 and C H202. in each buffer solution by using the volumes of each solution specified in the experiment protocol. Table view List view Initial Concentrations of Conjugates in Buffer Solutions [HC2H302] (M) [C2H3021(M) Solution 1: Final volume = 10 mL [HC2H302] = 0.50 M [NaC2H302] = 0 M Solution 4 Final volume = 10 mL [HC2H302] = (0.50 x 4) / 10 = 0.2 M [NaC2H302] = (0.50 x 6)/10 = 0.3 M Solution 1 0.50 0 Solution 2 0.40 0.10 Solution 2: Final volume = 10 mL [HC2H302] = (0.50 x 8) / 10 = 0.4 M [NaC2H302] = (0.50 x 2)/ 10 = 0.1 M Solution 5: Final volume = 10 mL [HC2H302] = (0.50 x 2)/10 = 0.1 M [NaC2H2O2) = (0.50 x 8) / 10 = 0.4 M 0.30 0.20 Solution 3 0.20 0.30 Solution 4 0.10 0.40 Solution 5 Solution 3: Final volume = 10 mL [HC2H302] = (0.50 x 6)/10 = 0.3 M [NaC2H302] = (0.50 x 4)/10 = 0.2 M Solution 6: Final volume = 10 mL [HC2H302] = 0 M [NaC2H302] = 0.50 M 0 0.50 Solution 6 2) Table view List view Difference between measured pH and calculated theoretical pH of solutions Measured pH Calculated pH ApH (do not include negative signi) Solution 1 2.51 2.52 0.01 Solution 2 4.15 4.15 0 Solution 3 4.54 4.57 0.03 Workings: Ka of acetic acid = 1.8 x 105 pka of acetic acid = 4.75 General formula for pH calculation of buffer: pH = pka + log (salt] / [acid)) where, (salt] = [NaC2H302) and (acid] = [HC2H302] Solution 1: Final volume = 10 mL [HC2H302] = 0.50 M [NaC2H302] = 0 M Since acetic acid is a weak acid, let us consider it produces a MH* ions on dissociation at equilibrium. Therefore, [H] = C2H302-) = a and [HC2H302] = 0.50-a Hence, Ka = 1.8 x 105 = a? / (0.50 - a) or, a = 3 x 103 M Hence, pH = -log (3 x 103) = 2.52 ApH = 2.52 - 2.51 = 0.01 Solution 4 Final volume = 10 mL [HC2H302] = (0.50 x 4) / 10 = 0.2 M [NaC2H2O2) = (0.50 X 6)/10 = 0.3 M pH = 4.75 + log (0.3/0.2) = 4.92 ApH = 4.92 - 4.89 = 0.036 Solution 4 4.89 4.92 0.036 Solution 5 5.33 5.35 0.02 Solution 6 9.27 9.22 0.05 Solution 5 Final volume = 10 mL [HC2H302] = (0.50 x 2)/10 = 0.1 M [NaC2H2O2) = (0.50 x 8) / 10 = 0.4 M pH = 4.75 + log (0.4/0.1) = 5.35 ApH = 5.35 - 5.33 = 0.02 Solution 2: Final volume = 10 mL [HC2H302] = (0.50 x 8) / 10 = 0.4 M [NaC2H302] = (0.50 x 2) / 10 = 0.1 M pH = 4.75 + log (0.1/0.4) = 4.15 ApH = 4.15 - 4.15 = 0 Solution 3 Final volume = 10 mL [HC2H302] = (0.50 x 6)/10 = 0.3 M [NaC2H302] = (0.50 x 4)/10 = 0.2 M pH = 4.75 + log (0.2 /0.3) = 4.57 ApH = 4.57 - 4.54 = 0.03 Solution 6 Final volume = 10 mL [HC2H302] = 0 M [NaC2H302] = 0.50 M Since sodium acetate is a salt of weak acid and strong base it dissociates in water to produce NaOH which undergoes further dissociation to give OH". Let us consider it produces b MOH ions Therefore, at equilibrium (NaOH] = [HC2H2O2] = b M and (C2H302] = (0.50-b) M Now, Kb = Kw/ka = 10-14 / 1.8 x 105 = 5.55 x 10-10 Hence, Kb = 5.55 x 10-10 = b2 /(0.50 - b) or, b = 1.67 x 10-5 M Hence, pOH = -log (1.67 x 10) = 4.78 pH = 14 - 4.78 = 9.22 ApH = 9.27 - 9.22 = 0.05 3) pH after addition of 3M HCI: Buffer Capacity Solution 1 O drops of HCI Solution 2 O drops of HCI Solution 3 O drops of HCI Solution 4 10 drops of HCI Solution 5 10 drops of HCI Solution 6 O drops of HCI pH after addition of 3M NaOH: Buffer Capacity Solution 1 O drops of NaOH Solution 2 5 drops of NaOH Solution 3 10 drops of NaOH Solution 4 5 drops of NaOH Solution 5 O drops of NaOH Solution 6 O drops of NaOH