A buffer system contains 0.30 M NH4+ and 0.15 M NH3. pKa of NH4+ is 9.25. How many moles of NaOH must be added to 1.00 L of this solution to increase the pH to 9.25? Hint: What is the (NH3) / (NH4+) when the pH=pKa?
If you want pH = pKa then you want (NH3) = (NH4^+). You get that from
pH = pKa + log (NH3)/(NH4^+)
9.25 = 9.25 + (NH3)/(NH4^+)
and that means (NH3) = (NH4^+)
......NH4^+ + OH^- ==> NH3 + H2O
I.....0.3..............0.15
add............x.............
C......-x.....-x.......+x
E...0.3-x......0......0.15+x
So you want 0.3-x = 0.15+x
Solve for x (in mols/L) and convert to mols in the 1 L.
When the pH of a buffer solution equals the pKa of the weak acid, the ratio of the concentration of the weak base (NH3) to the concentration of the weak acid (NH4+) is 1:1.
In this case, the pKa of NH4+ is given as 9.25. Therefore, when the pH is also 9.25, the concentration of NH3 will be equal to the concentration of NH4+.
To calculate the moles of NaOH needed to increase the pH to 9.25, we need to determine the number of moles of NH4+ and NH3 in the solution, and then find out the difference in moles required to achieve a 1:1 ratio.
Given:
Concentration of NH4+ (C1) = 0.30 M
Concentration of NH3 (C2) = 0.15 M
Volume of the solution (V) = 1.00 L
First, calculate the moles of NH4+ and NH3 in the solution:
Moles of NH4+ = C1 * V
Moles of NH4+ = 0.30 * 1.00 = 0.30 mol
Moles of NH3 = C2 * V
Moles of NH3 = 0.15 * 1.00 = 0.15 mol
Since the ratio of NH3 to NH4+ is 1:1 when the pH = pKa, we can conclude that the moles of NaOH required is the difference between the moles of NH4+ and NH3:
Moles of NaOH = Moles of NH4+ - Moles of NH3
Moles of NaOH = 0.30 - 0.15 = 0.15 mol
Therefore, 0.15 moles of NaOH must be added to 1.00 L of the solution to increase the pH to 9.25.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of the acid and its conjugate base. In this case, the acid is NH4+ (ammonium ion) and its conjugate base is NH3 (ammonia).
The Henderson-Hasselbalch equation is expressed as follows:
pH = pKa + log([A-] / [HA])
Before we can use this equation, we need to determine the initial ratio of [NH3] / [NH4+]. According to the problem, the buffer solution contains 0.30 M NH4+ and 0.15 M NH3. Therefore, the initial ratio is:
[NH3] / [NH4+] = 0.15 / 0.30 = 0.5
Now, we are asked to calculate how many moles of NaOH must be added to the solution to increase the pH to 9.25. At pH = pKa, the ratio of [NH3] / [NH4+] is 1. By adding NaOH, we will be converting NH4+ to NH3, increasing the [NH3] and subsequently altering the ratio.
Let's substitute the values into the Henderson-Hasselbalch equation:
9.25 = 9.25 + log([NH3] / [NH4+])
To find the new ratio [NH3] / [NH4+], we solve the equation:
log([NH3] / [NH4+]) = 9.25 - 9.25
log([NH3] / [NH4+]) = 0
[NH3] / [NH4+] = 1
Therefore, we need to add enough NaOH to convert NH4+ to NH3 until the ratio [NH3] / [NH4+] becomes 1.
Since we have 1.00 L of solution, we can use the initial ratio (0.5) to determine the initial moles of NH4+:
Moles of NH4+ = 0.5 * 0.30 mol/L * 1.00 L = 0.15 mol
To achieve a ratio of 1 between NH3 and NH4+, we need to convert all the NH4+ to NH3. This means we need to add:
0.15 mol of NaOH
Therefore, the number of moles of NaOH that must be added to 1.00 L of the buffer solution to increase the pH to 9.25 is 0.15 mol.