In the unshaded portions of the following table, supply the calculated molaritie
ID: 937880 • Letter: I
Question
In the unshaded portions of the following table, supply the calculated molarities of the indicated species in the solution mixtures. Assume the solutions are at 25 degrees Celsius. Note that in the rows labeled "Before Reaction," you must supply the appropriate concentrations after dilution occurs in the reaction mixture, but before any reaction takes place.
C. pH of Buffer Solutions:
1. Preparation of Buffer Solution: Weigh 3.3 g of solid sodium acetate trihydrate, NaC2H3O2 * H2O, into a clean 150 or 250 mL beaker. Using graduated cylinders, add 46 mL of deionized water and 4.0 mL of 6.0 M HC2H3O2. Mix thoroughly. Measure 19.0 mL of this into each of two clean beakers. Label as "A" and "B". Measure the pH of this buffer solution using the remaining 10 mL.
2. Addition of Acid or Base to Buffer:
a. To the buffer in beaker "A", add 1.0 mL of 3.0 M HCl. Mix thoroughly and read the pH of the resulting mixture.
b. To the buffer in beaker "B,", add 1.0 mL of 3.0 M NAOH. Mix thoroughly and read the pH of the resulting mixture.
3. pH of Deionized water: Obtain 19 mL of deionized water in each of two beakers. Read the pH of the deionized water. (NOTE: It will not be a pH of 7.00.)
4. Addition of Acid or Base to Deionized Water:
a. To one beaker of deionized water, add 1.0 mL of 3.0 M HCl. Mix thoroughly and read the pH of the resulting mixture.
C pH of Buffer Solutions In the unshaded portions of the following table, supply the calculated molarities of the indicated species in the solution mixtures. Assume the solutions are at 25°C Note that in the rows labeled Before Reaction," you must supply the appropri- ate concentrations after dilution occurs in the reaction mixture, but before any eaction takes place. Calculated Molarities in Mixtures HC NOH C.1 Buffer At Equilibrium C.2a Buffer+HC Before Reaction Ar Equilibrium C.2b Buffer +NaOH Before Reaction At Equilibrium C.3 Water Ar Equilibrium C.4a Water +HCI Before Reaction At Equilibrium C.4b Water+NaOH Before Reaction At EquilibriumExplanation / Answer
C.1. Buffer:
Given the mass of sodium acetate trihydrate(NaC2H3O2 * 3H2O) = 3.3 g
molecular mass of sodium acetate trihydrate(NaC2H3O2 * 3H2O) = 136.07 g/mol
Hence moles of sodium acetate trihydrate(NaC2H3O2 * 3H2O) = 3.3 g / 136.07 g/mol = 0.02425 mol
Total volume of the buffer, Vt = 46 mL+4 mL = 50 mL = 0.050 L
Hence [C2H3O2-] = 0.02425 mol / 0.050L = 0.485 M (answer)
Concentration of HC2H3O2 = 6.0 M
volume of HC2H3O2 = 4 mL = 0.004 L
Hence moles of HC2H3O2 in the buffer = MxV = 6.0 M x 0.004 L = 0.024 mol
Hence concentration of HC2H3O2 in the buffer, [HC2H3O2] = 0.024 mol / 0.050L = 0.480 M (answer)
for HC2H3O2, pKa = 4.76
Applying Hendersen equation
pH = pKa + log [C2H3O2-] / [HC2H3O2] = 4.76 + log(0.485 / 0.480) = 4.764
=> [H+] = 10-4.764 = 1.72x10-5 M(answer)
2(a): Before reaction: moles of C2H3O2- in 19 mL = 0.485M x 0.019 = 0.009215 mol
Total volume = 19+1 = 20 mL = 0.020 L
[C2H3O2-] = 0.009215 mol / 0.020 = 0.461 M(answer)
moles of HC2H3O2 in 19 mL = 0.480M x 0.019 = 0.00912 mol
Total volume = 19+1 = 20 mL = 0.020 L
[HC2H3O2] = 0.00912 mol / 0.020 = 0.456 M(answer)
moels of HCl = MxV = 3.0 M x 0.001L = 0.003 mol
[HCl] = 0.003 mol / 0.020 L = 0.15 M (answer)
After reaction: 0.003 mol of HCl will react with 0.003 mol of C2H3O2- to form 0.003 mol of 0.003 mol HC2H3O2.
Hence moles of C2H3O2- = 0.009215 - 0.003 = 0.006215 mol
[C2H3O2-] = 0.006215 mol / 0.020 = 0.311 M (answer)
moles of HC2H3O2 = 0.00912 + 0.003 = 0.01212mol
[HC2H3O2] = 0.01212 mol / 0.020 = 0.606 M (answer)
pH = 4.76 + log (0.311 / 0.606) = 4.470
=> [H+] = 3.39x10-5 M (answer)