How many mmol ( millimoles) of HCl must be added to 130.0 mL of a 0.40M solution of methyl amine (pKb = 3.36) to give a buffer having a pH of 11.59?
Thank you for your help!
This looks almost like the KOH/acetic acid above. If you still have trouble show what you've done and where you're stuck as well as what you don't understand and I can help you through it.
the correct answer is 5.2 mmol; however, I get lost in the calculations.
It seems easy to solve: I have pH, pKb, and HH eq, but then I get stuck to derive x correctly
Did you convert pKb given to pKa?
mmols CH3NH2 - 0.40 x 130 mL = 52
..........CH3NH2 + HCl ==> CH3NH3^+
I..........52.......0........0
add.................x.............
C.........-x.......-x........x
E.........52-x......0........x
11.59 = 10.64 + log (52-x)/(x)
8.91 = (52-x)/x
8.91x = 52-x
9.91x = 52
x = 5.24 which rounds to 5.2
Thank you! I finally have solved it
To determine the number of mmol of HCl needed to create a buffer, we need to understand the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of its acid and conjugate base components.
The Henderson-Hasselbalch equation is given as:
pH = pKa + log ([A-]/[HA])
In this case, we have methyl amine (CH3NH2) as the conjugate base (A-) and its corresponding acid will be CH3NH3+ (methyl ammonium ion). To find the number of mmol of HCl needed, we need to apply the Henderson-Hasselbalch equation. However, first, we need to determine the pKa of the methyl ammonium ion.
The pKa can be calculated from the pKb using the relation:
pKa + pKb = 14
So, pKa = 14 - pKb
pKa = 14 - 3.36
pKa = 10.64
Now we can apply the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Rearranging the equation, we get:
log ([A-]/[HA]) = pH - pKa
[A-]/[HA] = 10^(pH - pKa)
Substituting the given pH and pKa:
[A-]/[HA] = 10^(11.59 - 10.64)
[A-]/[HA] = 10^0.95
Now we can find the ratio of [A-] to [HA]. Since HCl will react with CH3NH2 (A-) to form CH3NH3+ (HA), we need to know the volume and concentration of CH3NH2.
The number of moles of CH3NH2 in 130.0 mL of 0.40M solution can be calculated using the formula:
moles = concentration × volume (in liters)
Converting the volume to liters:
130.0 mL / 1000 mL/L = 0.130 L
Moles of CH3NH2 = 0.40 M × 0.130 L
Moles of CH3NH2 = 0.052 mol
Since the ratio of [A-] to [HA] is 10^0.95, the number of mmol of HCl needed will be:
mol HCl = 0.052 mol × 10^0.95
mol HCl = 0.052 mol × 9.54993
mol HCl = 0.49796 mol
Converting mol to mmol (millimoles):
mmol HCl = 0.49796 mol × 1000
mmol HCl = 497.96 mmol
Therefore, you will need to add approximately 497.96 mmol of HCl to the 130.0 mL of the 0.40M solution of methyl amine to create a buffer with a pH of 11.59.