Part A)
find the pH of 9x10^-2 M of hypochlorus acid
Part B)
find pH of 7.8x10^-3 M of phenol
Part C)
find pH of 9.2x10^-2 of hydroxylamine
Part A) find the pH of 9x10^-2 M of hypochlorus acid Part B) find pH of...
Part A A 0.150 M weak acid solution has a pH of 2.97. Find Ka for the acid. Part B Find the percent ionization of a 0.195 M HC2H3O2 solution. (The value of Ka for HC2H3O2 is 1.8×10−5.) Part C Find the pH of a 0.0191 M solution of hypochlorous acid. (The value of Ka for hypochlorous acid is 2.9×10−8.) Part D Find the pH of a 0.014 M solution of HF. (The value of Ka for HF is 3.5×10−4.) Part...
A. Find "[OH-]= ____ M" for 5.0x10-2 M NaBrO. B. Find the pH for part A. C. Find [OH-] = ___ M" for a mixture that is 0.12 M in NaNO2 and 0.15 M in Ca(NO2)2. D. Find the pH for part C.
Determine the pH of each of the following solutions. Express your answer using two decimal places. A. 9.90×10−2 M hypochlorous acid. B. 7.9×10−3 M phenol. C. 9.6×10−2 M hydroxylamine.
***CHEMISTRY: ACIDS/BASES PROBLEM****
Determine the pH of each of the following solutions. ***Please
find the Ka and Kb values by doing a google search!!**
Part B 9.0x10-3 M phenol. Express your answer using two decimal places. pH4.26 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining Part C 9.0x102 M hydroxylamine. Express your answer using two decimal places. pH- Submit My Answers Give Up
Part C Find the pH of a 0.140 M solution of a weak monoprotic acid having Ka= 1.9×10−3. Part D Find the percent ionization of a 0.140 M solution of a weak monoprotic acid having KKa= 1.9×10−3.
Hypochlorous acid HClO Ka:3.0 * 10-8 Phenol C6H5OH (or HC6H5O) Ka:1.3 * 10-10 Hydroxylamine HONH2 Kb:1.1 * 10-8 Determine the pH of each of the following solutions (Ka and Kb values are given) all at 25C 1) 9.00×10−2 M hypochlorous acid. 2)7.9×10−3 M phenol. 3)9.0×10−2 M hydroxylamine.
The pH of a 0.25 M weak acid is 2.036. What is the Ka of the acid? HA(aq) + H2O (l) ↔ H3O+(aq) + A-(aq) 3.5 X 10-4 9.2X 10-3 7.0 X10-5 5.3 X 10-2 6.3 X 10-7 Explanation: [ H3O+] = antilog (-pH) =10-pH =10-2.306=9.2 X10-3 M [ H3O+] = [A-]=9.2 X10-3 M [HA] = 0.25 M- 0.0092 M=0.241 M Ka = [9.2 X10-3 M] [9.2 X10-3 M]/0.241 =3.52X10-4 I don't get why 10^-2.036= 9.2x10^3
Determine the pH of each of the following solutions. (a) 0.213 M boric acid (weak acid with Ka = 5.8e-10). (b) 0.394 M phenol (weak acid with Ka = 1.3e-10). (c) 0.869 M pyridine (weak base with Kb = 1.7e-09).
Part E Find the pH of a 0.140 M solution of a weak monoprotic acid having Ka= 0.12. Part F Find the percent ionization of a 0.140 M solution of a weak monoprotic acid having Ka= 0.12.
1) Find the pH of 0.200 M HF acid solution 2) Find the pH of a and [H3O+] of 0.100 M benzoic acid solution Explain how you got answer!