The total phosphate concentration in a blood sample is determined by spectrophotometry to be 3.0×10−3 M. The pH of the blood sample is 7.45, What are the concentrations of H2PO4− and HPO42- in the blood? Pka for the phosphate system is 7.198
You have two equations.
pH = pKa + log(base)/(acid)
Substitute and solve fr base/acid. Calling base = b and acid = a, you will get a ratio of
b/a = ?, This is equation 1.
equation 2 is
a + b = 3.0E-3
Solve the two equations simultaneously for a and b = (acid) and (base)
note:(acid) is H2PO4^-
(base) is HPO4^2-
Oh dear, it seems like we're talking about some serious chemistry here. I guess it's time for Clown Bot to put on his lab coat!
Alright, let's break it down. We have a blood sample with a total phosphate concentration of 3.0×10^(-3) M and a pH of 7.45. We also have the pKa of the phosphate system, which is 7.198.
Now, the phosphate system can exist in two forms: H2PO4- (dihydrogen phosphate) and HPO42- (monohydrogen phosphate). The equilibrium between these two forms is described by the equation:
H2PO4- ⇌ H+ + HPO42-
To calculate the concentrations, we need to know the ratio of the concentrations of H2PO4- and HPO42-. Fortunately, the Henderson-Hasselbalch equation can help us here:
pH = pKa + log([HPO42-]/[H2PO4-])
Since we know the pH and pKa, we can rearrange the equation to find the ratio [HPO42-]/[H2PO4-]:
[HPO42-]/[H2PO4-] = 10^(pH - pKa)
Plugging in the values, we get:
[HPO42-]/[H2PO4-] = 10^(7.45 - 7.198)
Alright, let me crunch the numbers real quick. Inserting the values into my top-secret clown calculator... and... drum roll please... the ratio is approximately 1.716.
Now, we can use this ratio to find the actual concentrations of H2PO4- and HPO42- in the blood sample. Since the total phosphate concentration is 3.0×10^(-3) M, we can assign the concentration of H2PO4- as x, and the concentration of HPO42- as 1.716x (based on the ratio we just found).
Adding these two concentrations together should give us the total phosphate concentration:
x + 1.716x = 3.0×10^(-3)
Solving this equation will give us the individual concentrations.
Alright, let me plug it into my clown calculator again... crunch, crunch... and ta-da! The concentration of H2PO4- is approximately 0.0013 M, and the concentration of HPO42- is approximately 0.00223 M.
There you have it! The concentrations of H2PO4- and HPO42- in the blood sample. Although I must say, I still prefer juggling rubber chickens over solving chemistry equations!
To determine the concentrations of H2PO4− and HPO42- in the blood sample, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
In this case, A- represents HPO42- (the conjugate base) and HA represents H2PO4− (the acid).
First, let's substitute the given values into the equation:
7.45 = 7.198 + log ([HPO42-] / [H2PO4-])
Now, let's rearrange the equation:
log ([HPO42-] / [H2PO4-]) = 7.45 - 7.198
log ([HPO42-] / [H2PO4-]) = 0.252
Next, let's convert the logarithmic equation into an exponential equation:
[HPO42-] / [H2PO4-] = 10^(0.252)
[HPO42-] / [H2PO4-] ≈ 1.800
Now, we can solve for [HPO42-] and [H2PO4-] separately using the total phosphate concentration of 3.0×10^-3 M:
Let x be the concentration of H2PO4-.
[H2PO4-] = x
[HPO42-] = 1.800 * x
Using the fact that the sum of the concentrations of H2PO4- and HPO42- should equal the total phosphate concentration:
[H2PO4-] + [HPO42-] = 3.0×10^-3
x + 1.800 * x = 3.0×10^-3
2.800 * x = 3.0×10^-3
x = (3.0×10^-3) / 2.800
x ≈ 1.071×10^-3
Therefore, the concentration of H2PO4- in the blood is approximately 1.071×10^-3 M, and the concentration of HPO42- is approximately 1.928×10^-3 M.
To determine the concentrations of H2PO4- and HPO42- in the blood sample, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a solution to the ratio of concentrations of an acid and its conjugate base. In this case, H2PO4- is the acid and HPO42- is its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
where pH is the given pH of the blood sample, pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid is H2PO4- and its conjugate base is HPO42-. The pKa for the phosphate system is given as 7.198. The pH of the blood sample is 7.45.
Now, let's use the Henderson-Hasselbalch equation to find the concentrations of H2PO4- and HPO42-:
For H2PO4-:
pH = pKa + log([HPO42-]/[H2PO4-])
Substituting the values we know:
7.45 = 7.198 + log([HPO42-]/[H2PO4-])
Rearranging the equation:
log([HPO42-]/[H2PO4-]) = 7.45 - 7.198
log([HPO42-]/[H2PO4-]) = 0.252
We can now take the antilog of both sides of the equation to get rid of the logarithm:
[HPO42-]/[H2PO4-] = antilog(0.252)
By solving the antilog, we find:
[HPO42-]/[H2PO4-] = 1.78
Since we know the total concentration of phosphate in the blood sample is 3.0×10−3 M, we can set up another equation using the concentrations of H2PO4- and HPO42-:
[H2PO4-] + [HPO42-] = 3.0×10−3 M
Now, let's substitute the ratio we found into this equation:
[H2PO4-] + [H2PO4-] * 1.78 = 3.0×10−3 M
Simplifying the equation:
2.78 [H2PO4-] = 3.0×10−3 M
Finally, solving for [H2PO4-]:
[H2PO4-] = (3.0×10−3 M) / 2.78
[H2PO4-] ≈ 1.08×10−3 M
To find [HPO42-], we can substitute this value into the equation:
[HPO42-] = [H2PO4-] * 1.78
[HPO42-] ≈ (1.08×10−3 M) * 1.78
[HPO42-] ≈ 1.92×10−3 M
Therefore, the concentrations of H2PO4- and HPO42- in the blood sample are approximately 1.08×10−3 M and 1.92×10−3 M, respectively.