Part A Part A Determine the [H3O+] of a 0.250 M formic acid solution. Express your answer using two significant digits. [H3O+] 🇧🇷 Part B Part B Determine the pH of this Formica solution
acid. p H Express your answer with two decimal places. pH=???
Part A:
HCOOH < -------------->
COO-(aq) + H3O+(aq)
I
0.250
0
VS
-X
+x
+x
AND
0.250-x
+x
+x
Ka = [H3O+][COO-] 🇧🇷 [HCOOH] = 1.8 × 10^-4
1.8 x 10^-4 = x^2 / (0.250-x)
x^2 + (1.8 x 10^-4)x – (4.5 x 10^-5) = 0
Solve the quadratic equation,
x = 0.0066
This way, [H3O+] =x=0.0066M
🇧🇷
Part B:
pH = -log[H3O+]
pH = -log(0.0066)
pH = 2.18
Dice, 0.250 M HCOOH (formic acid)
[H3O+]
Data data Concentration of formic acid HCOOH Equilibrium constant is given by 0.250 I 0.250 F 0.250-x Equilibrium constant is given by [H30][COO] HCOOH Substitutek for 1.8×10″, cool and H,O]for x and[HCOOH]at 0.125x [H30. ][C00] HCOOH 4 1.8×100.125-x x2+1.8x10x-1×10 0 x 0.0066 So the concentration is HO 0.0066 M
Calculate the pH of the solution. pH recording[H,o] Replace 0.0066 with HO pH-log(H,o) =-log(0.0066) – 2.18 The pH of the solution is therefore 2.18
Latest posts by Answer Prime (see all)