A buffer solution is made that is 0.363 M in HClO and 0.363 M in NaClO . (1) If
ID: 500714 • Letter: A
Question
A buffer solution is made that is 0.363 M in HClO and 0.363 M in NaClO .
(1) If Ka for HClO is 3.50×10-8 , what is the pH of the buffer solution?
(2) Write the net ionic equation for the reaction that occurs when 0.082 mol HCl is added to 1.00 L of the buffer solution.
Use H3O+ instead of H+ .
A buffer solution is made that is 0.388 M in HCN and 0.388 M in NaCN.
(1) If Ka for HCN is 4.00×10-10, what is the pH of the buffer solution?
(2) Write the net ionic equation for the reaction that occurs when 0.114 mol NaOH is added to 1.00 L of the buffer solution.
Please explain how you got this answer!
Explanation / Answer
Q1.
find pH of solution,
HClO and ClO- is present, so there is buffer formation
pH = pKa + log(ClO-/HClO)
pKa = -log(3.5*10^-8) = 7.45
[ClO-] = 0.363 [HClO] = 0.363
substitute in buffer equation
pH = 7.45+ log(0.363/0.363)
pH = 7.45
Q2.
the net ionic equation for 0.082 mol of HCl V = 1 L so
[HClO] = 0.363 + 0.082 = 0.445
[Cl-O] = 0.363 - 0.082 = 0.281
pH = pKa + log(ClO-/HClO)
recalculate pH
pH = 7.45 + log(0.281 /0.445 )
pH = 7.250
net ionic:
ClO-(aq) + H3O+(aq) --> HClO(aq) + H2O(l)
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