Asked by Oli
You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 10-4). A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 liter of buffered solution?
How does the henderson hasslebache equation help find the volume?
How does the henderson hasslebache equation help find the volume?
Answers
Answered by
DrBob222
pH = pKa + log [(base)/(acid)]
You know Ka. Calculate pKa.
NO2^- is the base.
HNO2 is the acid.
First, plug in pH = 3.00, the pH of the solution you want, and calculate the (base)/(acid) ratio.
That gives you (base) = factor x (acid).
Then you know mLbase*Mbase = mLacid*Macid AND you know mLacid + mL bse = 1000.
You solve for mLacid and mL base. If I didn't goof, it is something like 700 mL acid and 300 mL base.
You know Ka. Calculate pKa.
NO2^- is the base.
HNO2 is the acid.
First, plug in pH = 3.00, the pH of the solution you want, and calculate the (base)/(acid) ratio.
That gives you (base) = factor x (acid).
Then you know mLbase*Mbase = mLacid*Macid AND you know mLacid + mL bse = 1000.
You solve for mLacid and mL base. If I didn't goof, it is something like 700 mL acid and 300 mL base.
Answered by
DrBob222
Those volumes I quoted are not the exact volumes but they are close to the correct values.
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