Buffers Adding to a solution?
Answer:
Ka of HCN is 4 x 10^-10
pKa = - log 4 x 10^-10 = 9.40
For a buffer solution we can use Henderson- Hasselbalch equation :
pH = pK + log [ CN-] / [ HCN]
6.2 = 9.40 + log [ CN-] / 0.15
- 3.2 = log [CN- ]/ 0.15
10^-3.2 = [CN-] / 0.15
[CN-] = 9.46 x 10^-5 M
Molecular weight NaCN = 49 g/mol
Moles NaCN in 1 L = 9.46 x 10^-5
Moles NaCN in 350 mL = 350 x 9.46 x 10^-5 /1000 = 3.31 x 10^-5
49 g/mol x 3.31 x 10^-5 mol = 0.00162 g of NaCN
Look up the pKa of HCN.
Take the buffer equation
Plug in the pKa of HCN.
Plug in 6.2 for the pH
Plug in 0.15 for [HA]
Solve for the concentration [A(-)]
Your answer is [A(-)] x 0.35 L x 49 g / mol.
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