Question
1. Assume that 1.00 mol of HCOOH and 0.725 mol of HCOONa are diluted in an...
1. Assume that 1.00 mol of HCOOH and 0.725 mol of HCOONa are diluted in an aqueous solution of 1.00 L of solution:
a. Calculate the pH of the solution (Ka = 1.77 x 10-4)
b. Assume that 0.10 mol of a strong acid such as HCl is added to the previous HCOOH / HCOONa solution. Calculate the pH of the new solution.
Answers
1.00 1@ liven, ka=1.77x104 moles of H200nt = 1:00 mel moles of HCOONa = 0.725 mol. Use Henderson - Hasselbalch equation - pH = pka + log [ conjugate Base] [weak Acid] P!= -log ka + loa [HCOona] [4cook] = -log 1.77x104 + log 0.725 = 3677 +(-0.14) 1 pH = 3.63 Hence pH of solution is pH = 3.63 16 on addition of 0.10 mol of HCL HCOONa aunt HCleah HCOOH (aq) Before he Aditon tron 0.725 mol 0.10 mol 1.00 medl After Addition (HCl) 0.625 mol omol 1.10 mol I Can Befor Conc. Conc. Hence nce pHepka + log [conjugate Base] 19 [weak Acid J pH = -log ka + log II Hook] - - log 16.77X107 +log (1-10) = 3.77–0-2455 = -log 1.77x154 + log 10.625]= 3.5245 12 = 3:53) - pH of New solution after Addition 1.00 mol Hel is pH = 3.53
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