A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH of this buffer?

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

What is the pH of the buffer after the addition of 0.02 mol of HI?

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=5.05

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml