Answer:
1.- pH =3.61
2.-pH =3.53
Explanation:
In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,
pH = pKa + log [A⁻]/[HA]
where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].
In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.
First Part
pH = 3.40+ log {0.170 /0.105}
pH = 3.61
Second Part
# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012
# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)
# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol
# mol NaNO₂ remaining = (0.170 - 0.012) mol = 0.158
# mol HNO₂ produced = 0.012 mol
# mol HNO₂ initial = 0.105
# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol
Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.
pH = 3.40 + log {0.158/.0117}
pH = 3.53
Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.
Answer:
The answers are:
pH buffer= 3.61
pH buffer after the addition of HCl= 3.53
Explanation:
HNO₂ is a weak acid and NaNO₂ is the salt of a weak acid (the conjugated base is NO₂⁻). In order to calculate the pH of a buffer solution we use the Henderson-Hasselbach equation:
pH= pKa + log [tex]\frac{[salt]}{[acid]}[/tex]
For this buffer we have:
[salt]= [NaNO₂] = 0.170 M
[acid]= [HNO₂] = 0.105 M
pKa= 3.40
V= 1 L
Thus, we introduce the data in the equation and calculate the pH of the buffer solution:
pH= 3.40 + log (0.170 M/0.105 M) = 3.61
After the addition of HCl (a strong acid), the salt (NaNO₂) will react with HCl to produce the weak acid HNO₂. So, the concentration of the salt will decrease and the concentration of the acid will increase. We have to recalculate the concentrations of salt and acid as follows:
The moles of HCl added are:
HCl moles= 12.0 mol/L x 1 L/1000 ml x 1.00 ml = 0.012 moles
The total volume of the solution will be : 1 L + (1 ml x 1 L/1000 ml= 0.001 L)
Vt= 1 L + 0.001 L = 1.001 L
We have to reduce the total moles of NaNO₂ in 0.012 moles and to increase the total moles of HNO₂ in 0.012 moles.
moles NaNO₂ = initial moles NaNO₂ - HCl moles
= (0.170 mol/L x 1 L) - 0.012 mol
= 0.158 mol
[NaNO₂] = moles NaNO₂/Vt= 0.158 mol/1.001 L = 0.1578 M
moles HNO₂ = initial moles HNO₂ + HCl moles
= (0.105 mol/L x 1 L) + 0.012 mol
= 0.117 mol
[HNO₂] = moles HNO₂/Vt= 0.117 mol/1.001 L= 0.1168 M
Finally we introduce the concentrations in the Henderson- Hasselbach equation and calculate the pH after the addition of HCl:
pH= 3.40 + log (0.1578 M/0.1168 M) = 3.53
