Asked by Simpson
A buffer is prepared by dissolving .0250 mol of sodium nitrite, NaNO2, in 250.0 mL of .0410 M nitrous acid, HNO2. Assume no volume change after HNO2 is dissolved. Calculate the pH of this buffer.
My book does a poor job of explaining buffers and I'm really lost on the concept. Any directions and explanations would be appreciated.
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]/
My book does a poor job of explaining buffers and I'm really lost on the concept. Any directions and explanations would be appreciated.
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]/
Answers
Answered by
Josh
[HNO2] = .0410 M
[NaNO2] = .0250 mol / .250 L = .100 M
[H+] = Ka x [HA (weak acid)] / [A- (conj. base)]
Ka = 6.0 x 10^-4 (a given constant)
[H+] = (6.0 x 10^-4) x [.041]/[.100]
= 2.5 x 10^-4 M.
pH = -Log[H+] -- pH = -log[2.5 x 10^-4]
pH = 3.61
[NaNO2] = .0250 mol / .250 L = .100 M
[H+] = Ka x [HA (weak acid)] / [A- (conj. base)]
Ka = 6.0 x 10^-4 (a given constant)
[H+] = (6.0 x 10^-4) x [.041]/[.100]
= 2.5 x 10^-4 M.
pH = -Log[H+] -- pH = -log[2.5 x 10^-4]
pH = 3.61
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