Question

In: Chemistry

For 530.0 mL of a buffer solution that is 0.155 M in CH3CH2NH2 and 0.145 M...

For 530.0 mL of a buffer solution that is 0.155 M in CH3CH2NH2 and 0.145 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Expert Solution

V = 530 ml

M = 0.155 CH3CH2NH2

M = 0.145 CH3CH2NH3Cl

a) find pH initial

this is abuffer

its modeled with henderson haselback equation

pOH = pKb + log(HB+ / B)

Kb = 5.0*10^-4

pKb = -log(Kb) =-log(5.0*10^-4)= 3.3

pOH = 3.3 + log(0.145/0.155 )

pOH = 3.27

but we need pH so

pH = 14-pOH = 14-3.27 = 10.73

pH =10.73

b) after 0.02 mol of HCl

We will react base (neutralize it)

CH3CH2NH2 + HCl ---> H2O + CH3CH2NH3Cl

ratio is 1:1 so 1 mol of acid reacts with 1 mol of base to form salt + h2o

The moles before reaction

mol of CH3CH2NH2 = M*V = 0.155*¨0.53 = 0.08215

mol of CH3CH2NH3Cl = M*V = 0.145*0.53 = 0.07685

mol of HCl = 0.02

After reaction:

mol of CH3CH2NH2 =0.08215 - 0.02 = 0.06215

mol of CH3CH2NH3Cl = 0.07685 + 0.02 = 0.09685

mol of HCl = 0.02 - 0.2 = 0

there is no need to recalculate concnetrations since it is the same volume i.e. M = M and n = n since V = V

therefore

pOH = 3.3 + log(0.09685/0.06215 ) = 3.3 + 0.1926 = 3.49

pH =14 -pOH = 14-3.49 = 10.50

queen_honey_blossom answered 3 weeks ago

For 260.0 mL of a buffer solution that is 0.2869 M in CH3CH2NH2 and 0.2669 M...

For 260.0 mL of a buffer solution that is 0.2869 M in CH3CH2NH2 and 0.2669 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Kb=5.6⋅10−4).

For 220.0 mL of a buffer solution that is 0.2869 M in CH3CH2NH2 and 0.2624 M...

For 220.0 mL of a buffer solution that is 0.2869 M in CH3CH2NH2 and 0.2624 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Kb=5.6⋅10−4).

For 220.0 mL of a buffer solution that is 0.3028 M in CH3CH2NH2 and 0.2783 M...

For 220.0 mL of a buffer solution that is 0.3028 M in CH3CH2NH2 and 0.2783 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Kb=5.6⋅10−4).

For 200.0 mL of a buffer solution that is 0.2975 M in CH3CH2NH2 and 0.2730 M...

For 200.0 mL of a buffer solution that is 0.2975 M in CH3CH2NH2 and 0.2730 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Kb=6.46⋅10−4).

For 550.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.175 M...

For 550.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.175 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl. Express your answers using two decimal places separated by a comma.

PART 1: For 500.0 mL of a buffer solution that is 0.155 M in HC2H3O2 and...

PART 1: For 500.0 mL of a buffer solution that is 0.155 M in HC2H3O2 and 0.145 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.020 mol of HCl. ... PART 2: For 500.0 mL of a buffer solution that is 0.150 M in CH3CH2NH2 and 0.135 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

For 540.0mL of a buffer solution that is 0.165M in CH3CH2NH2 and 0.155M in CH3CH2NH3Cl, calculate...

For 540.0mL of a buffer solution that is 0.165M in CH3CH2NH2 and 0.155M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020mol of HCl.

If 100.mL of 0.035 M HCl solution is added to 220.mL of a buffer solution which...

If 100.mL of 0.035 M HCl solution is added to 220.mL of a buffer solution which is 0.20M in NH3 and 0.18M in NH4Cl, what will be the pH of the new solution? The Kb of NH3 is 1.8x10^-5. Please explain how the answer was obtained. Thank you!

11 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that...

11 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.08 M C2H3O2-. What is the pH of the resulting solution?

A buffer solution was prepared by mixing 300.0 mL of a 0.200 M solution of hydroxylamine...

A buffer solution was prepared by mixing 300.0 mL of a 0.200 M solution of hydroxylamine (OHNH2, Kb = 1.1 × 10-8) and 250.0 mL of a 0.300 M solution of its hydrochloride salt (OHNH3+ Cl- ). Calculate the pH of the solution after the addition of 1.08 g of solid NaOH (molar mass = 40.00 g/mol). Assume that there is no volume change upon the addition of solid.