[products] over [reactants]
To calculate the Ka value of the weak acid HA, you can use the pH of the solution and the formula for calculating the Ka. First, calculate the concentration of [H+], which is 10^(-pH). Then, use the expression for Ka: Ka = [H+][A-]/[HA], where [A-] and [HA] are assumed to be equal in a weak acid solution. Plug in the [H+] value you calculated and the initial concentration of HA to find Ka.
For a weak acid, HA...HA ==> H^+ + A^- Ka = [H+][A-]/[HA] Plug these values into the Ka equation. You also must know the [HA] that you start with. Solve for [H+] Take -log [H+] = pH
H2CO3---------- 2 H+ + (CO3)2-
The resulting equation may not be equivalent to the original equation because raising both sides of an equation to the same power is an operation that introduces extraneous solutions. This can occur when the original equation contains roots or fractional exponents. It's important to verify solutions to ensure they satisfy the original equation.
For a weak acid HA, Ka = [H+] [A-]/[HA] where Ka is the dissociation constant and the square brackets are the concentrations of the species shown in mol dm-3. Percentage ionization is given by PI = [H+] /[HA] x 100 Let the molarity of the final solution be M. Then we have, for the original solution,PI = [H+] /[HA] x 100 = [H+] x 100/0.2PI= 500 [H+]From the first equation, remembering that the concentrations of the two ions are equal,[H+] = sq rt(0.2 Ka)Thus , original PI = 500 x sq rt (0.2 Ka)For the final solution, PI = [new H+]/M x 100 and as before, [new H+] = sq rt (KaM)So final PI = sq rt (KaM)/M x 100= sqrt (Ka)/sqrtM x100We know this is twice the original PI sosqrt (Ka)/sqrtM x100 = 2 x 500 x sq rt (0.2 Ka)Solving this gives M = 0.05.This is a quarter of the original molarity, ( 0.05/0.2), so the final volume should be 4 times the original, i.e. 400 ml.
To calculate the Ka of an acid, you can use the equation Ka H3OA- / HA, where H3O is the concentration of hydronium ions, A- is the concentration of the conjugate base, and HA is the concentration of the acid. The Ka value represents the acid's strength in donating protons in a solution.
The determination of Ka for a weak acid in the lab involves measuring the concentration of the acid and its conjugate base at equilibrium, and using this information to calculate the acid dissociation constant (Ka) using the equation Ka HA-/HA.
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I would take the equation to calculate the new amount, and solve it for the original amount.
The relationship between Ka and Kb values is that they are related by the equation Kw Ka Kb, where Kw is the ion product of water. If you know the Kb value, you can determine the Ka value by rearranging the equation to solve for Ka.
β = 2.3C {Ka[H3O+] / (Ka+[H3O+])2}
balance your chemical reaction equation then calculate moles, then calculate weight.
a = KB/KA
maa ka bhosada
Extraneous solution
To find the Ka value for HF, we can use the equation for the dissociation of HF into H⁺ and F⁻ ions: HF ⇌ H⁺ + F⁻. First, calculate the initial concentration of HF using the given pH. Then, determine the concentrations of H⁺ and F⁻ ions at equilibrium. Finally, use these concentrations to calculate the Ka value using the equation Ka = [H⁺][F⁻] / [HF].
Its an equation you can use to find the pH of a solution. it is.... --- pH = pKa + log (Base/Acid) --- these may help too Ka = 10^-pKa Kw = Ka*Kb