A buffer solution of volume 50.0 mL contains 12.5 mmol HNO2 and 9.25 mmol KNO2.
1. What is the pH of this solution?
2. If 1.00 mmol of Ca(OH)2 is added to the solution, what will be the new pH?
3. If 2.00 mL of 2.00M HCl is added to the original solution, what will be the new pH?
If you could explain it step by step I'd really appreciate it!
1. Use the HH equation.
pH = pKa + log (KNO2)/(HNO2)
2. Without values of pKa any thing I calculate won't agree with your calculations. Here is what you do.
.......HNO2 + OH^- ==> NO2^- + H2O
I......12.5....0........9.25
add............2............
C......-2.....-2........+2
E......10.5....0........+2
Substitute the E line into the HH equation and solve for the new pH.
3. Follow the steps in 3 but
.......NO2^- + H^+ ==> HNO2
I......9.25....0........12.5
add............4............
C......-4.....-4........+4
E......etc.
Sure! I'd be happy to explain it step by step.
To answer these questions, we'll need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
where pH is the measure of acidity or alkalinity, pKa is the logarithmic acid dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
Let's go through each question one by one:
1. What is the pH of this solution?
In this case, we have the concentration of both the acid (HNO2) and its corresponding conjugate base (NO2-). Given that the volume of the solution is 50.0 mL, we can convert the given concentrations from mmol/mL (millimoles per milliliter) to M (molarity) by dividing by 1000:
[HNO2] = 12.5 mmol / 50.0 mL = 0.25 M
[KNO2] = 9.25 mmol / 50.0 mL = 0.185 M
The pKa value for HNO2 is typically around 3.15 at 25°C. Now, using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
= 3.15 + log (0.185/0.25)
≈ 3.15 + log (0.74)
≈ 3.15 - 0.13 (approximating log(0.74) to -0.13)
≈ 3.02
Therefore, the pH of the buffer solution is approximately 3.02.
2. If 1.00 mmol of Ca(OH)2 is added to the solution, what will be the new pH?
When calcium hydroxide (Ca(OH)2) reacts with HNO2, it will form calcium nitrate [Ca(NO3)2] and water (H2O). However, the reaction between Ca(OH)2 and HNO2 is not a typical acid-base reaction because Ca(OH)2 is a strong base, and HNO2 is a weak acid. Consequently, buffering capacity of the solution will be reduced, resulting in a significant change in pH.
To calculate the new pH, we need to determine how much of the HNO2 and NO2- will be consumed. Since Ca(OH)2 is a strong base, it will react completely with HNO2 according to the stoichiometry:
Ca(OH)2 + 2HNO2 -> Ca(NO2)2 + 2H2O
So, for every 1 mmol of Ca(OH)2 added, 2 mmol of HNO2 will be consumed. Thus, the remaining HNO2 concentration will be:
[HNO2] = 12.5 mmol - 2 mmol = 10.5 mmol / 50.0 mL = 0.21 M
Similarly, the remaining NO2- concentration will be:
[KNO2] = 9.25 mmol / 50.0 mL = 0.185 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
= 3.15 + log (0.185/0.21)
≈ 3.15 + log (0.881)
≈ 3.15 - 0.054 (approximating log(0.881) to -0.054)
≈ 3.10
Therefore, the new pH of the solution, after adding 1.00 mmol of Ca(OH)2, is approximately 3.10.
3. If 2.00 mL of 2.00M HCl is added to the original solution, what will be the new pH?
When HCl, a strong acid, is added to the buffer solution, it will react with the weak base, NO2-. HCl will convert NO2- into its conjugate acid, HNO2.
To calculate the new pH, we need to determine how much of NO2- will be converted into HNO2. Using the equation:
[HNO2] = 0.25 M + (2.00 mL / 50.0 mL) * 2.00 M
= 0.25 M + 0.08 M
= 0.33 M
Now, the new [NO2-] concentration will be:
[KNO2] = 9.25 mmol - (2.00 mL / 50.0 mL) * 2.00 M
= 9.25 mmol - 0.08 mmol
= 9.17 mmol / 50.0 mL
= 0.183 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
= 3.15 + log (0.183/0.33)
≈ 3.15 + log (0.554)
≈ 3.15 - 0.255 (approximating log(0.554) to -0.255)
≈ 2.90
Therefore, the new pH of the solution, after adding 2.00 mL of 2.00 M HCl, is approximately 2.90.