Assume that the normal blood buffer contains 0.00080 M carbonic acid and 0.0085 M hydrogen carbonate, the pKa = 6.35 for carbonic acid and the volume of blood in the body is 7.00 L. The blood pH, due to disruption is now 7.20. What is the ratio of [HCO3−]/[H2CO3] now that the blood has been challenged? How many moles of hydrogen carbonate must be added to the blood to bring the carbonic acid/hydrogen carbonate ratio back to a normal pH = 7.4? How many milliliters of 0.10 M hydrogen carbonate must be added to the blood?
To find the ratio of [HCO3−]/[H2CO3] in the blood after the disruption, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([HCO3−]/[H2CO3])
Given:
pH = 7.20
pKa = 6.35
Let's solve for [HCO3−]/[H2CO3]:
7.20 = 6.35 + log([HCO3−]/[H2CO3])
Subtracting 6.35 from both sides:
0.85 = log([HCO3−]/[H2CO3])
Using the definition of logarithms, we can rewrite this equation as:
10^0.85 = [HCO3−]/[H2CO3]
Calculating 10^0.85:
7.079 = [HCO3−]/[H2CO3]
So, the ratio of [HCO3−]/[H2CO3] in the blood after the disruption is approximately 7.079.
To determine the number of moles of hydrogen carbonate (HCO3−) that must be added to the blood to restore the carbonic acid/hydrogen carbonate ratio to a normal pH of 7.4, we can use the equation derived from the Henderson-Hasselbalch equation:
[HCO3−]/[H2CO3] = 10^(pH - pKa)
Given:
pH = 7.4
pKa = 6.35
[HCO3−]/[H2CO3] = 7.079 (from the previous calculation)
Let's solve for [HCO3−]/[H2CO3]:
7.079 = 10^(7.4 - 6.35)
Calculating 10^(7.4 - 6.35):
7.079 = 10^1.05
7.079 = 10.599
Now, to determine the number of moles of hydrogen carbonate (HCO3−) needed to restore the ratio, we need to compare it to the initial concentration of carbonic acid (H2CO3) in the blood.
[H2CO3] = (0.00080 M) × (7.00 L) = 0.00560 moles
So, the number of moles of hydrogen carbonate (HCO3−) needed is:
0.00560 moles × (10.599 - 1) = 0.05624 moles
To calculate the volume of 0.10 M hydrogen carbonate solution needed to contribute 0.05624 moles of HCO3−, we can rearrange the equation:
moles = concentration × volume
0.05624 moles = (0.10 M) × volume
Solving for volume:
volume = 0.05624 moles / 0.10 M
volume ≈ 0.5624 L
Converting liters to milliliters:
volume ≈ 562.4 mL
Therefore, approximately 562.4 milliliters of 0.10 M hydrogen carbonate solution must be added to the blood.
To find the ratio of [HCO3−]/[H2CO3] after the blood has been challenged, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([HCO3−]/[H2CO3])
Given that the blood pH is 7.20, the pKa is 6.35, and we want to find the ratio [HCO3−]/[H2CO3], we can rearrange the equation:
[HCO3−]/[H2CO3] = 10^(pH - pKa)
Plugging in the values:
[HCO3−]/[H2CO3] = 10^(7.20 - 6.35)
[HCO3−]/[H2CO3] = 10^0.85
[HCO3−]/[H2CO3] ≈ 7.08
So, the ratio of [HCO3−]/[H2CO3] now that the blood has been challenged is approximately 7.08.
To find the moles of hydrogen carbonate that must be added to bring the ratio back to a normal pH of 7.4, we can use the equation:
[HCO3−]/[H2CO3] = 10^(pH - pKa)
Given that the new pH is 7.4 and the pKa is 6.35, rearranging the equation:
[HCO3−]/[H2CO3] = 10^(7.4 - 6.35)
Plugging in the values:
[HCO3−]/[H2CO3] = 10^1.05
[HCO3−]/[H2CO3] = 10.65
To bring the ratio back to the normal value of 10.65, we need to increase the concentration of HCO3− without changing the concentration of H2CO3. This means we should add more HCO3− to the blood.
To find the moles of hydrogen carbonate to be added, we can use the formula:
moles = concentration * volume
Given the concentration of 0.0085 M hydrogen carbonate and the volume of blood being 7.00 L:
moles = 0.0085 M * 7.00 L
moles ≈ 0.0595 mol
So, approximately 0.0595 moles of hydrogen carbonate must be added to bring the carbonic acid/hydrogen carbonate ratio back to a normal pH of 7.4.
To find the milliliters of 0.10 M hydrogen carbonate to be added, we can use the formula:
moles = concentration * volume
Given the concentration of 0.10 M hydrogen carbonate, we need to rearrange the formula to find the volume:
volume = moles / concentration
Plugging in the values:
volume = 0.0595 mol / 0.10 M
volume = 0.595 L
Since we want the volume in milliliters, we can convert the result:
volume = 0.595 L * 1000 mL/L
volume = 595 mL
So, approximately 595 milliliters of 0.10 M hydrogen carbonate must be added to the blood.
First I would check to see what the pH is at these concentrations and I like to work in millimoles so
base = 0.0085 x 7.00L = 59.5 mmols
acid = 0.008 x 7.00L = 5.6 mmols.
pH = 6.35 + log(59.5/5.6) = 7.38
I look at the second part of the problem by asking how much acid must I add to the "normal" blood to make it pH of 7.20 and that will be how much bicarbonate I must add to get it back from 7.20 to normal 7.38.
........HCO3^- + H^+ ==> H2CO3
I.......59.5..............5.6
add..............x..............
C.......-x......-x........+x
E......59.5-x....0.......5.6+x
Plug into the HH equation and solve for x, then add to 5.6 and subtract from 59.5
I like to check to see that this actually produces the desired pH. Of course if this is how much acid must be added to produce that "off beat" pH of 7.20 it will be the bicarbonate that must be added to get it back to normal.
You can find the ratio of the base/acid at pH 7.20 from the HH equation. Just plug in the values.
E........................59.5+x