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Buffer sln
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It is easy but I cannot solve or do buffer questions/slns.. Please help.
Describe the preparation of 500 ml of 1 M phosphate
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Answers (5)
a + b = 0,5 then a=b= 0,25 mol 0.25 mol / 0,5 l = 0,5 M NaH2PO4 (30 gr & 0,5 M) 0,25 / 0,5 = 0,5 M Na2HPO4 (35,5 gr & 0,5 M) Is it ok now?
OMG!! log 1 = 0 so b/a = 1 I am deeply embarrassed :( why ý write log 0 = 0 :O
I saw this : when pKa = pH, there is equal concentration of acid and its conjugate base. So is what ý did before true ?
Ok there is a problem again. My lecturer said take the values showed on lecture. Values are so: pKa1= 2,14 pKa2= 7,2 pKa3= 12,4 when I take pKa2= 7,2 then log b/a = 0 then b=0 a= 0,5 . Don't we use any base? 7.2 = pKa2 + log b/a 7.2 = 7.20 + log b/a log
I did this: as you told, this is wrong :( H3PO4 ↔ H2PO4 + H+ ↔ HPO4 + H+ ↔ PO4 + H+ pH: 7,2 Prepare 500 ml of 1 M NaH2PO4. 500ml = 0,5 liter 0,5 l * 1M = 0,5 mol NaH2PO4 1 mol NaH2PO4 = 120 gr, so; 0,5 mol NaH2PO4 = 60 gr Solve 60 gr NaH2PO4 in 500
