The weak acid HA has a Ka of 4.5×10−6. If a 1.4 M solution of the acid is prepared, what is the pH of the solution? The equilibrium expression is:
HA(aq)+H2O(l)⇋H3O+(aq)+A−(aq)
We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
The weak acid HA has a Ka of 4.5×10−6. If a 1.4 M solution of the...
What is the pH of a 0.44 M solution of a weak acid HA, with a Ka of 3.19×10−12? The equilibrium expression is: HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)
The weak acid HIO has a Ka of 2.0×10−11. If a 1.7 M solution of the acid is prepared, what is the pH of the solution? The equilibrium expression is: HIO(aq)+H2O(l)⇋H3O+(aq)+IO−(aq) Report your answer with two significant figures.
The weak acid C,H,OH has a Ka of 1.6 x 10-10. If a 1.4 M solution of the acid is prepared, what is the pH of the solution? The equilibrium expression is: C6H,OH(aq) + H2O(l) — H,0+(aq) + CH 0-(aq) • Round your answer to two decimal places. Provide your answer below: pH=
A weak acid, HA, is a monoprotic acid. A solution that is 0.250 M in HA has a pH of 1.890 at 25°C. HA(aq) + H2O(l) ⇄ H3O+(aq) + A-(aq) What is the acid-ionization constant, Ka, for this acid? What is the degree of ionization of the acid in this solution? Ka = Degree of ionization =
The weak acid HA has a Ka of 1.80×10−5. If a 1.9 M solution of the acid is prepared, what is the pH of the solution?
show all work Consider that 20.0 mL of 0.10 M HA (an arbitrary weak acid, Ka= 2.5 × 10−6) is titrated with 0.10 M NaOH solution. The ionization of HA in water occurs as the following. HA (aq) + H2O(l) ⇌ A (aq) + H3O (aq) The neutralization reaction between HA and NaOH can be expresses as the following. HA (aq) + NaOH (aq) NaA (aq) + H2O (l) Answer the following questions. A) What will be the initial...
ASAP! What is Ka for the weak acid, HA, if a 0.020 M solution of the acid has a pH of 3.29 at 25ºC? a. 5.1 × 10-2 b. 6.9 × 10-2 c. 2.6 × 10-4 d. 1.3 × 10-5 e. 1.0 × 10-6 What is the conjugate acid of H2PO4–(aq)? a. H3O+ b. H3PO4 c. HPO42– e. PO43–
The weak acid HA has a Ka of 6.64×10−5. If a 1.9 M solution of the acid is prepared, what is the pH of the solution? a)pH=3.0 b)pH=5.2 c)pH=6.3 d)pH=7.5
What is the pH of a 6.85 × 10−3 M weak acid solution, HA, if Ka = 4.5 × 10−6? Group of answer choices 1.2 4.8 9.1 3.8 6.5 What is the pOH of 4.50 × 10−4M HBr? Group of answer choices 10.7 6.7 1.7 12.3 3.3 What is the pH of a 9.67 × 10−3M solution of NaOH? Group of answer choices 13.0 4.6 12.0 9.4 2.0 A 6.5 × 10-2 M solution of a weak acid, HA, has...
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