1) Part A a solution that is 0.175 M in HC2H3O2 and 0.120 M in KC2H3O2 Express y

Question

1)

Part A

a solution that is 0.175 M in HC2H3O2 and 0.120 M in KC2H3O2

Express your answer using two decimal places.

Part B

a solution that is 0.195 M in CH3NH2 and 0.110 M in CH3NH3Br

Express your answer using two decimal places.

2)

Part A

Write an equation showing how this buffer neutralizes added acid (HNO3).

Express your answer as a chemical equation.

Part B

Write an equation showing how this buffer neutralizes added base (CsOH).

Express your answer as a chemical equation.

3)

Calculate the ratio of CH3NH2 to CH3NH3Cl required to create a buffer with pH = 10.20.

Express your answer using two significant figures.

Explanation / Answer

Solution :-

1)

Part A

a solution that is 0.175 M in HC2H3O2 and 0.120 M in KC2H3O2

solution :- using the Henderson equation we can calculate the pH of the buffer

pH= pka + log [base/acid]

pH= 4.74 + log [0.120 / 0.175]

pH=     4.58

Part B

a solution that is 0.195 M in CH3NH2 and 0.110 M in CH3NH3Br

Solution :- Ka for the CH3NH3^+ = 1*10^-14 / kb

                                                      = 1*10^-14 / 4.4*10^-4

                                                       = 2.27*10^-11

Pka = -log ka

Pka = 2.27*10^-11

Pka = 10.64

Now lets calculate the pH

pH= pka + log [base /acid]

pH = 10.64 + log [0.195/0.110]

pH= 10.89

2)

Part A

Write an equation showing how this buffer neutralizes added acid (HNO3).

Solution :- When HNO3 is added to the buffer then it reacts with conjugate base as follows

CH3COO^- + HNO3 ------ > CH3COOH + NO3^-

Part B

Write an equation showing how this buffer neutralizes added base (CsOH).

Solution :-

When CsOH is added to buffer then it will react with acid as follows

CsOH + CH3COOH ------ > CH3COOCs + H2O

Or we can write it as

OH- + CH3COOH   -------- > CH3COO^- + H2O

3)

Calculate the ratio of CH3NH2 to CH3NH3Cl required to create a buffer with pH = 10.20

Solution :-

Pka of the CH3NH3^+ = 10.64

Using the Henderson equation lets calculate the ratio of the base to acid to get pH 10.20

pH= pka + log [base/ acid]

10.20 = 10.64 + log [CH3NH2/CH3NH3Cl]

10.20 – 10.64 = log [CH3NH2/CH3NH3Cl]

-0.44 = log [CH3NH2/CH3NH3Cl]

Antilog (-0.44) = [CH3NH2/CH3NH3Cl]

0.36 = [CH3NH2/CH3NH3Cl]

So the ratio is 0.36

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