Question

1. For 280. 0 mL of a buffer solution that is 0. 225 M in HCHO2 and 0. 300 M in KCHO2, calculate the initial pH and the final pH after adding 0. 028 mol of NaOH. ( Ka(HCHO2)=1. 8×10−4. ) Express your answers to two decimal places. Enter your answers numerically separated by a comma.

2. For 280. 0 mL of a buffer solution that is 0. 295 M in CH3CH2NH2 and 0. 225 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0. 028 mol of NaOH. ( Kb(CH3CH2NH2)=5. 6×10−4. )

Express your answers to two decimal places. Enter your answers numerically separated by a comma.

Answer 1

1.For the buffer solution containing HCHO2 and KCHO2:

First, we can calculate the moles of HCHO2 and KCHO2 present in the solution:

moles of HCHO2 = (0.225 M) x (0.2800 L) = 0.063 moles

moles of KCHO2 = (0.300 M) x (0.2800 L) = 0.084 moles

Since NaOH is a strong base, it will react completely with the weak acid, HCHO2, to form the conjugate base, CHO2-. We can use the balanced chemical equation to determine the moles of HCHO2 that will react with NaOH:

HCHO2 + NaOH -> H2O + NaCHO2

1 mole of HCHO2 reacts with 1 mole of NaOH. Therefore, since we are adding 0.028 mol of NaOH, 0.028 mol of HCHO2 will react.

The amount of HCHO2 and CHO2- in the buffer solution after the reaction can be calculated as follows:

moles of HCHO2 = 0.063 - 0.028 = 0.035 moles

moles of CHO2- = 0.084 + 0.028 = 0.112 moles

Next, we can calculate the concentration of HCHO2 and CHO2- in the buffer solution after the reaction:

[ HCHO2 ] = moles of HCHO2 / volume of solution = 0.035 moles / 0.2800 L = 0.125 M

[ CHO2- ] = moles of CHO2- / volume of solution = 0.112 moles / 0.2800 L = 0.400 M

Using the Henderson-Hasselbalch equation, we can calculate the initial pH of the buffer solution:

pH = pKa + log([ CHO2- ] / [ HCHO2 ])

pH = -log(1.8x10^-4) + log(0.400 / 0.125)

pH = 3.91

Finally, we can calculate the final pH after the addition of NaOH. The NaOH reacts with HCHO2 to form CHO2-, which will increase the concentration of the conjugate base and decrease the concentration of the weak acid. The new concentrations of HCHO2 and CHO2- are:

[ HCHO2 ] = 0.035 moles / 0.2800 L = 0.125 M

[ CHO2- ] = 0.140 moles / 0.2800 L = 0.500 M

Using the Henderson-Hasselbalch equation again, we can calculate the final pH of the solution:

pH = pKa + log([ CHO2- ] / [ HCHO2 ])

pH = -log(1.8x10^-4) + log(0.500 / 0.125)

pH = 4.32

Therefore, the initial pH of the buffer solution is 3.91, and the final pH after the addition of NaOH is 4.32.

2.For the buffer solution containing CH3CH2NH2 and CH3CH2NH3Cl:

First, we can calculate the moles of CH3CH2NH2 and CH3CH2NH3Cl present in the solution:

moles of CH3CH2NH2 = (0.295 M) x (0.2800 L) = 0.0826 moles

moles of CH3CH2NH3Cl = (0.225 M) x (0.2800 L

Regenerate response