A 500-mL buffer was prepared and it is 0.300 M in nitrous acid (HNO2 , Ka = 4.0 x 10-4) and 0.200 M in sodium nitrite, NaNO2. What will be the pH of this solution after adding 1.00 mL of 6.00 M HCl?
millimols HNO2 = M x mL = 0.300 x 500 = 150
mmols NaNO2 = 0.2 x 500 = 100
mmols HCl added = 6 x 1 = 6
----------------------------
.........NO2^- + H^+ ==> HNO2
I.......100......0........150
add..............6..................
C........-6.....-6........+6
E........94......0.......156
Now substitute the E line into the Henderson-Hasselbalch equation and solve for pH.
To determine the pH of the solution after adding 1.00 mL of 6.00 M HCl, we need to consider the reaction between HNO2 (nitrous acid) and HCl (hydrochloric acid) and its effect on the buffer.
First, let's find the initial moles of HNO2 and NaNO2 in the 500 mL buffer solution.
Initial moles of HNO2 = initial concentration of HNO2 × volume of solution
= 0.300 M × 500 mL
= 0.150 moles
Initial moles of NaNO2 = initial concentration of NaNO2 × volume of solution
= 0.200 M × 500 mL
= 0.100 moles
After adding 1.00 mL of 6.00 M HCl, we assume complete dissociation of HCl to form H+ ions and Cl- ions. So, there will be an additional 6.00 mmol of H+ ions in the solution.
Now, let's consider the reaction between HNO2 and HCl:
HNO2 + HCl -> NO2- + H2O + Cl-
Using the Henderson-Hasselbalch equation, we can calculate the ratio of HNO2 to NO2-:
pH = pKa + log([NO2-]/[HNO2])
The pKa value for nitrous acid (HNO2) is given as 4.0 x 10^-4.
Considering the initial moles of HNO2 and NaNO2, we find the ratio:
[NO2-]/[HNO2] = (moles of NaNO2 + moles of H+)/(moles of HNO2 - moles of H+)
Substituting the values:
[NO2-]/[HNO2] = (0.100 moles + 0.006 moles)/(0.150 moles - 0.006 moles)
Simplifying,
[NO2-]/[HNO2] = 0.671
Now, substituting the ratio [NO2-]/[HNO2] and the pKa value into the Henderson-Hasselbalch equation:
pH = 4.0 x 10^-4 + log(0.671)
Using logarithmic properties, we can calculate the log value and then the pH:
pH = 4.0 x 10^-4 + (-0.175)
pH = 3.825
Therefore, the pH of the solution after adding 1.00 mL of 6.00 M HCl will be approximately 3.825.