5)
use:
pH = pKa + log {[conjugate base]/[acid]}
3.95 = pKa + log {15/10}
pKa = 3.774
use:
pKa = -log Ka
3.774 = -log Ka
Ka = 1.7*10^-4
Answer: d
5. A buffer solution contains 10.0 mmol of formic acid (HCOOH) and 15.0 mmol of formate...
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23. A 1.0 L buffer solution is prepared with 0.25 M formic acid (HCOOH) and 0.50 M sodium formate (HCOONa). HCOOH(aq) + H2O(aq) – HCOO (aq) + H30+(aq) pKa = 3.74 What is the pH of this buffer solution? (A) 3.74 (B) 4.04 (C) 4.50 (D) 4.20
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a 100 ml solution of 0.250 M formic acid (HCOOH) was titrated to its equivalence point with 50 mL of sodium hydroxide. The complete molecular equation for the reaction is shown below HCOOH (aq) + NaOH (aq)---------> HCOONa (aq) +H20 (l) Ka of HCOOH= 1.7 x 10 ^-4 calculate the pH at the equivalence point
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