How many moles of conjugate base (Na2HPO4) are present in 40mL of .010M buffer at pH 7.5 vs at pH 6?
Buffer was made up of Na2HPO4 and NaH2PO4.
pH=pKa +log[A/HA]; pKa=7.2.... When I did the calculations I got the same amount as if it were in a 0.1M buffer at the same pH's so I think I'm doing something wrong.
I showed you how to do this for pH 7.5. You do it the same way for pH 6 and the answers are not the same. In instead of us working this again, why not show your work and let us find the error.
@DrBob222 oh, so instead of solving for change in pH we stop after a+b=4mmols. so it'll be different when it goes to .01M.
But 7.5=7.2+logb/a; b=2a a+b=.004mols; 3a=.004 a=.00376 b=.000237. So that would be my final answer for pH 7.5 and I do the same for pH 6?
And, the whole pH 6 thing still doesn't make sense. I just took the absolute value and got a change in pH of .0095.
I just responded to your first post down below about the pH 6 thing. The absolute value is not the way to go.
To calculate the number of moles of the conjugate base (Na2HPO4) present in the buffer, you need to take into account the volume of the buffer and the concentration of the buffer components.
Firstly, let's determine the concentration of Na2HPO4 and NaH2PO4 in the buffer solution. You mentioned that the buffer was prepared with Na2HPO4 and NaH2PO4, and the concentration of the buffer is given as 0.010M.
Since Na2HPO4 is the conjugate base, we can assume that it is present in excess and can be considered the main component. Therefore, the concentration of Na2HPO4 is also 0.010M.
Next, we need to consider the pH of the buffer solution. The Henderson-Hasselbalch equation allows us to calculate the ratio of conjugate base (A = Na2HPO4) to acid (HA = NaH2PO4) at a given pH:
pH = pKa + log([A]/[HA])
For pH 7.5:
7.5 = 7.2 + log([A]/[HA])
To calculate the ratio [A]/[HA], we can rearrange the equation:
log([A]/[HA]) = 7.5 - 7.2
log([A]/[HA]) = 0.3
Next, we need to convert the logarithm to an exponential form:
[A]/[HA] = 10^0.3
So, [A]/[HA] = 2
Since the concentration of Na2HPO4 (A) is 0.010M, the concentration of NaH2PO4 (HA) can be calculated as follows:
[HA] = [A]/2 = 0.010 / 2 = 0.005 M
Now, to determine the number of moles of Na2HPO4 in the buffer solution:
40 mL is equal to 0.040 L (converting from milliliters to liters).
For pH 7.5:
moles of Na2HPO4 = concentration of Na2HPO4 * volume of buffer solution
= 0.010 M * 0.040 L
= 0.0004 moles
Similarly, for pH 6:
pH = pKa + log([A]/[HA])
6 = 7.2 + log([A]/[HA])
log([A]/[HA]) = 6 - 7.2
log([A]/[HA]) = -1.2
[A]/[HA] = 10^(-1.2)
[A]/[HA] = 0.063
So, the concentration of NaH2PO4 (HA) is 0.063 * 0.010 M = 0.00063 M.
moles of Na2HPO4 = concentration of Na2HPO4 * volume of buffer solution
= 0.010 M * 0.040 L
= 0.0004 moles
Therefore, regardless of the pH (7.5 or 6), the number of moles of Na2HPO4 present in 40 mL of the buffer remains the same, which is 0.0004 moles.
If the calculated moles are the same for both pH values as you mentioned, then your calculations are correct. It appears that the change in pH does not affect the number of moles of Na2HPO4.