Respuesta :
Answer:
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
The pH of buffer solution of sodium propionate and propionic acid is 5.05; the addition of 0.02 mol NaOH has pH 5.13, and addition of 0.02 mol HI has pH 4.97.
What is an acid dissociation constant?
The acid dissociation constant is the pH at which the acid compound is dissociated into its constituent ions.
The buffer of propionic acid and sodium propionate has an acid dissociation constant ([tex]K_a[/tex]) of 4.989.
- The concentration of compounds in the buffer is given as:
[tex]\rm Concentration=\dfrac{moles}{Volume\;(L)}[/tex]
The volume of the buffer solution is 1.2 L.
The concentration of propionic acid is:
[tex]\rm [C_2H_5COOH]=\dfrac{0.18}{1.2}\\C_2H_5COOH=0.15\;M[/tex]
The concentration of salt, sodium propionate is:
[tex]\rm [C_2H_5COONa]=\dfrac{0.26}{1.2}\\ C_2H_5COONa=0.217\;M[/tex]
The pH of the buffer solution can be given as:
[tex]\rm pH=pK_a+ln\dfrac{salt}{acid}[/tex]
Substituting the values for pH of buffer:
[tex]\rm pH=4.989+ln\dfrac{0.217}{0.15}\\ pH=5.05[/tex]
The pH of the buffer solution is 5.05.
- With the addition of 0.02 mol, NaOH consumes 0.02 mol acid in the buffer solution resulting in salt formation.
The remaining acid in the buffer is:
[tex]\rm Acid=0.18-0.02\;mol\\Acid=0.16\;mol[/tex]
The moles of salt in the solution will be:
[tex]\rm Salt=0.26+0.02\;mol\\Salt=0.28\;mol[/tex]
The concentration of acid in the solution are:
[tex]\rm Acid=\dfrac{0.16}{1.2} \\Acid=0.133\;M[/tex]
The concentration of salt in the solution are:
[tex]\rm Salt=\dfrac{0.28}{1.2}\\ Salt=0.233\;M[/tex]
The pH of the solution will be:
[tex]\rm pH=4.89+ln\dfrac{0.233}{0.133}\\ pH=5.13[/tex]
The pH of the buffer with the addition of 0.02 mol NaOH is 5.13.
- The addition of 0.02 mol HI increases the acid concentration while decreasing the salt concentration.
The concentration of acid in the buffer will be:
[tex]\rm Acid=\dfrac{0.18+0.02}{1.2}\\ Acid=0.167\;M[/tex]
The concentration of salt in the buffer will be:
[tex]\rm Salt=\dfrac{x0.26-0.02}{1.2}\\ Salt=0.20\;M[/tex]
The pH of the buffer solution is given as:
[tex]\rm pH=4.89+ln\dfrac{0.20}{0.167}\\ pH=4.97[/tex]
The pH of the buffer with the addition of 0.02 mol HI is 4.97.
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