A beaker with 1.40×102mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.10mL of a 0.370M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
base = b
acid = a
First you must determine the (acid) and (base). You do that by solving two equations simultaneously.
b+a = 0.1M is equation 1
pH = pKa + log (acid)/(base) and substitute like this from the problem
5.00 = 4.76 + log (b/a) is equation 2.
Solve the two for a and b. These aren't the exact numbers but they're close.
a = about 0.04
b = about 0.06
millimols acid = 0.04*140 = about 5.6
mmols base = 0.06* 140 = 8.4
mmols HCl added = about 2.6
........b + H^+ ==> Hb
I......8.4..0.......5.6
add........2.6................
C.....-2.6.-2.6......+2.6
E.....5.8...0.......8.2
Substitute the E line into the HH equation and solve for pH.
Post your work if you get stuck.
Check this work carefully. I did it in a hurry.
To determine how much the pH will change after adding HCl solution to the acetic acid buffer, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the buffer solution
pKa = the dissociation constant of the acid
[A-] = the concentration of the conjugate base
[HA] = the concentration of the acid
In this case, we are adding HCl solution to the buffer, so the concentration of the acid ([HA]) will increase. However, the concentration of the conjugate base ([A-]) will also increase due to the reaction between HCl and the acetic acid in the buffer.
To calculate the new concentrations [A-] and [HA], we can use the formula:
[A-] = [initial A-] + [HCl]
[HA] = [initial HA] - [HCl]
Given data:
Initial [A-] concentration = 0.100 M (total molarity of acid and conjugate base)
Initial [HA] concentration = 0.100 M (total molarity of acid and conjugate base)
Volume of HCl solution added = 7.10 mL
Concentration of HCl solution = 0.370 M
Using the volume and concentration information, we can calculate the moles of HCl added:
moles of HCl added = volume of HCl solution added * concentration of HCl solution
moles of HCl added = 7.10 mL * 0.370 M
Now, we can calculate the new concentrations of [A-] and [HA]:
[A-] = 0.100 M + (moles of HCl added / total volume)
[HA] = 0.100 M - (moles of HCl added / total volume)
Next, we can substitute the new concentrations into the Henderson-Hasselbalch equation:
new pH = pKa + log([A-]/[HA])
Given:
pKa = 4.760
Calculate:
new pH = 4.760 + log([A-]/[HA])
Finally, we can calculate the new pH by substituting the calculated values:
new pH = 4.760 + log([A-]/[HA])
You can use a scientific calculator or logarithm table to find the logarithm value and obtain the new pH.