QUESTION 1 2 points Save Answer Consider a buffer solution made of 0.205 M formi

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QUESTION 1 2 points Save Answer Consider a buffer solution made of 0.205 M formic acid, HCOOH and 0.155 M sodium formate, HCOONa. Ka(HCOOH) = 1.8x10-4 After addition of 0.050 moles of NaOH to 1.0 L of this buffer, the pH becomesith blank. Show the mumber only, Report with 2 digits after the decimal point. QUESTION 2 2 points Save Answer A buffer is 0.275 M in aqueous ammonia (NH3) and 0.310 M in ammonium nitrate, NH4NO3. Kb(NH3) 1.8x10 The pH of this buffer is [Xin the bank; show the number only. Report with 2 digits after the decimal point Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers Save and Submit

Explanation / Answer

1)

mol of NaOH added = 0.05 mol

HCOOH will react with OH- to form HCOO-

Before Reaction:
mol of HCOO- = 0.155 M *1.0 L
mol of HCOO- = 0.155 mol

mol of HCOOH = 0.205 M *1.0 L
mol of HCOOH = 0.205 mol

after reaction,
mol of HCOO- = mol present initially + mol added
mol of HCOO- = (0.155 + 0.05) mol
mol of HCOO- = 0.205 mol

mol of HCOOH = mol present initially - mol added
mol of HCOOH = (0.205 - 0.05) mol
mol of HCOOH = 0.155 mol


Ka = 1.8*10^-4

pKa = - log (Ka)
= - log(1.8*10^-4)
= 3.745

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.745+ log {0.205/0.155}
= 3.866

Answer: 3.866

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