Thursday

July 31, 2014

July 31, 2014

Posted by **Natalie** on Monday, November 11, 2013 at 11:05pm.

- Chemistry -
**DrBob222**, Tuesday, November 12, 2013 at 12:10amFirst 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

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