What is the Ka for an acid HA, if the equilibrium concentrations are [HA]=0.822 M, [H3O+]=1.15×10−4 M, and [A−]=1.15×10−4 M?
What is the Ka for an acid HA, if the equilibrium concentrations are [HA]=0.822 M, [H3O+]=1.15×10−4...
What is the Ka of a weak acid (HA) if equilibrium concentrations are [H3O+] = [A-] = 6.1 x 10-4 M , [HA] = 0.25 M
The equilibrium concentrations of the reactants and products are [HA] = 0.260 M [H ] = 3.00 × 10–4 M [A–] = 3.00 × 10–4 M Calculate the Ka value for the acid HA.
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)
A diprotic acid, H,A, has acid dissociation constants of Ka molar concentrations of H,A, HA-, and A2- at equilibrium for each of the solutions 1.42 x 10-4 and Ka2 = 4.07 x 1012. Calculate the pH and = A 0.210 M solution of H,A H2A] = pH HA- A2- М М A 0.210 M solution of NaHA HA pH= М
A monoprotic acid, HA, is dissolved in water: HA <------> H+ +A- The equilibrium concentrations of the reactants and products are [HA] = 0.280 M [H ] = 2.00 × 10–4 M [A–] = 2.00 × 10–4 M. Calculate the value of pKa for the acid HA.
A 0.050 M solution of a weak monoprotic acid, HA, has [H3O+] = 3.8 x 10-5. What is the value of Ka for this acid? 2.6 x 10-11 2.9 x 10-8 7.5 x 10-3 7.0 x 10-8 7.0 x 10-7
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–
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)
What are the equilibrium concentrations of all the solute species in a 0.99 M solution of propanoic acid, HC3H5O2? (a) [H3O+], M; (b) [OH-], M; (c) [CH3CH2COOH], M; (d) What is the pH of the solution? For CH3CH2COOH, Ka = 1.34 x 10-5.
Determine the [H3O+] of a 0.220 M solution of formic acid (Ka=1.8×10−4).