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.
