Using stocks of 0.1 M KH2PO4 and 0.1 M K2HPO4 how much 0.1 M K2HPO4 is needed to prepare 20 mL of a buffer solution at pH 7.2. The pKa of phosphate is 6.82.
I know to solve for the ratio using henderson hasselbach but im lots after that!
5.2=6.82+log(A/HA)
1.0024HA=A
I think you need the total concentration of the buffer; i.e., you want 20 mL of what? 0.05M, 0.01M, what M. Here is a problem just like that which I worked earlier last week.
http://www.jiskha.com/display.cgi?id=1470250931
To prepare a buffer solution at pH 7.2 using the stocks of 0.1 M KH2PO4 and 0.1 M K2HPO4, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, A- represents the conjugate base (K2HPO4) and HA represents the acid (KH2PO4).
You are given the pH (7.2) and the pKa of phosphate (6.82). To find the ratio of [A-] to [HA], you can rearrange the equation:
7.2 = 6.82 + log([A-]/[HA])
Now, solve for ([A-]/[HA]):
0.38 = log([A-]/[HA])
To convert from logarithmic form to exponential form, you can rewrite the equation as:
10^(0.38) = [A-]/[HA]
Now, calculate 10^(0.38):
[A-]/[HA] ≈ 2.38
This means that the ratio of [A-] to [HA] should be close to 2.38.
Next, you need to determine how much of the 0.1 M K2HPO4 is needed to prepare a 20 mL buffer solution. Since you know the ratio of [A-] to [HA], you can set up the following equation:
([A-]/[HA]) = (x/0.1)
Where x represents the amount of 0.1 M K2HPO4 needed.
Substituting the ratio value:
2.38 = (x/0.1)
Now, solve for x:
x = 0.1 * 2.38
x ≈ 0.238
Therefore, you would need approximately 0.238 mL of the 0.1 M K2HPO4 stock solution to prepare a 20 mL buffer solution at pH 7.2.