Question

Find the pH of a solution prepared by adding 1.0 L of 0.10 M butyric acid to 0.50 L of 0.10 M NaOH. (Ka = 1.5 x 10-5)

A. 12.69

B. 7.00

C. 2.41

D. 4.82

E. 13.35

Answer 1

Solution : To solve this problem let us first calculate the initial No. of moles of both butyric acid and Naou. Molarity of butyple acid = 0.10 M. Volume of butyvie acid = il * No. of moles of butyrie acid = Molaniby * volume of solution = 0.1X1 - 0.molej Initial No. of moles of butyric acid = 0.1 moles Now molarity of Naon = 0.10m. volume of Naon = 0.5L E a No. of moles of Naon = Molarity & volume of solution = 0.100.5 = 0.05 moles DEUTSC E . Initial No. of moles of Noon = 0.05 moles 1 - 0 OS

Now, to solve for pht, we need BCA Table F.e., Before - After- Before - Change - Aftov. & Naon a Cuacuacu.coona + H₂O CH₂ CH, CH,cool °(Butyric Aciol) oil 0.05 -0.05 -0.05 +0:05 0.05 0.1-0.05 -0.05 to calculate the ple, we us Hendersen - Hanselbach equation bka + loal No. of moles of Conjugate base 1. ph- pray No of moles of Acid. But pka = -log ka = -log (1,5X10-5) = 4.82 :: pl = 4.82 + log (Novof males of company, Cu, C00Na1 moles of CU CU ₂ Cu₂coou :: PH: 4:82 + 10g (0:0) = 4.82.710911) = 4:8240 = 4.82 pH = 4.82