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What is the pka of .012M solution and a pH of 2.31?
HA ==> H+ + A-Ka = [H+][A-][HA] and from pH = 2.31, calculated [H+] = 4.89x10^-3 M Ka = (4.89x10^-3)(4.89x10^-3)/0.012 Ka = 1.99x10^-3 pKa = 2.70
How can you tell how strong acids and alkalis are?
Look it up in a table of acid constants (proteolytic equilibrium constants) or Ka-table. weak: CH3COOH, acetic acid Ka=1.7*10-5 , very weak: HCN, Hydrogen cyanide Ka = 4.9*10-10 strong: HCl, Hydrogen chloride Ka >> 1.0
What is the ka of the first ionization of h3po4 if you initially had 0.58 m of phosphoric acid and your pH at equilibrium is 3?
To find the Ka of the first ionization of H3PO4, you need to set up an ICE table and use the formula for Ka. Given that the pH at equilibrium is 3, you can calculate the concentration of H+ ions at equilibrium. Then, use this concentration to calculate the concentration of the other species in the reaction and plug them into the Ka expression to find the Ka value.
What is the concentration of acetic acid and the pH in a solution that is 0.25 ionized?
pka of acetic acid = 4.7 ka = 10-4.7 ka = 2.0 * 10-5 ka = [H][A]/[HA] 2.0 * 10-5 = [H][A]/0.02 3.99 * 10-7 = [H][A] [H] = 6.3 * 10-4 (square root of above number as [A] and [H] are assumed the same) pH = -log(6.3 * 10-4) pH = 3.2
Determine the pH of a 5 M HA solution at 25 degrees Ka 7.1 x 10-4?
Given that Ka = 7.1 x 10-4 pKa = -log[Ka] = 3.15 (2 s.f) pH = 1/2pka - 1/2log[A-] = 1/2(3.15) - 1/2log[5] = 1.575 - (-0.3495) pH = 1.93 (2 s.f.) So the pH of HA = 1.93
Formula for calculation pH and pOH?
pH + pOH =14
What is the pka of .012M solution and a pH of 2.31?
HA ==> H+ + A-Ka = [H+][A-][HA] and from pH = 2.31, calculated [H+] = 4.89x10^-3 M Ka = (4.89x10^-3)(4.89x10^-3)/0.012 Ka = 1.99x10^-3 pKa = 2.70
Busbar fault level calculation?
what do you mean by 50 KA for 1 sec
What is the pka of a .012M solution and a pH of 2.31?
HA ==> H+ + A-Ka = [H+][A-][HA] and from pH = 2.31, calculated [H+] = 4.89x10^-3 M Ka = (4.89x10^-3)(4.89x10^-3)/0.012 Ka = 1.99x10^-3 pKa = 2.70
Calculate the pH of 0.2 M HNO2 solution ( Ka = 4.5 * 10 -5 )?
pH= -log = 1.59
What is Henderson-Hasselbalch equation?
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
How do you determining pH from weak acids?
H2S >> H+ + HS- Ka = [H+][HS-]/[H2S] Ka = 10^-7.0
0.15M solution of weak acid HA has pHof 5.35 What is Ka?
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.
How can you tell how strong acids and alkalis are?
Look it up in a table of acid constants (proteolytic equilibrium constants) or Ka-table. weak: CH3COOH, acetic acid Ka=1.7*10-5 , very weak: HCN, Hydrogen cyanide Ka = 4.9*10-10 strong: HCl, Hydrogen chloride Ka >> 1.0
What is the ka of the first ionization of h3po4 if you initially had 0.58 m of phosphoric acid and your pH at equilibrium is 3?
To find the Ka of the first ionization of H3PO4, you need to set up an ICE table and use the formula for Ka. Given that the pH at equilibrium is 3, you can calculate the concentration of H+ ions at equilibrium. Then, use this concentration to calculate the concentration of the other species in the reaction and plug them into the Ka expression to find the Ka value.
What is the concentration of acetic acid and the pH in a solution that is 0.25 ionized?
pka of acetic acid = 4.7 ka = 10-4.7 ka = 2.0 * 10-5 ka = [H][A]/[HA] 2.0 * 10-5 = [H][A]/0.02 3.99 * 10-7 = [H][A] [H] = 6.3 * 10-4 (square root of above number as [A] and [H] are assumed the same) pH = -log(6.3 * 10-4) pH = 3.2
Determine the pH of a 5 M HA solution at 25 degrees Ka 7.1 x 10-4?
Given that Ka = 7.1 x 10-4 pKa = -log[Ka] = 3.15 (2 s.f) pH = 1/2pka - 1/2log[A-] = 1/2(3.15) - 1/2log[5] = 1.575 - (-0.3495) pH = 1.93 (2 s.f.) So the pH of HA = 1.93
