Acids and bases

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aqa chemistry
Anushka John
Flashcards by Anushka John, updated more than 1 year ago
Anushka John
Created by Anushka John over 7 years ago
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Question Answer
Define a Bronsted-Lowry acid A substance that can donate a proton
Define Bronsted-Lowry base A substance that can accept a proton
What is a conjugate base? What is left over after the acid has donated its proton
Calculating pH equation pH = -log [H+]
The concentration of H+ ions in a monoprotoic strong acid will be the _____ as ________ of the acid the same as the concentration of acid (as strong acids completely dissociate)
What is the pH for 0.1M HCl acid? pH = -log [0.1] = 1.00 This is because the conc. of H+ ions is the same as the conc. of the monoprotoic acid
Equation for finding the conc. of H+ from pH (STRONG ACID) [H+]= 10^(-pH)
What is the concentration of HCl with a pH of 1.35? [H+] = 10^(-1.35) = 0.045 M
In all aqueous solutions and pure water, which equilibrium occurs? H2O <---> H+ + OH-
Equation for the IONIC PRODUCT OF WATER Kw = [H+]
At room temperature, what is the value of Kw for all aqueous solutions 1 x 10^(-14)
How do we find the pH pf water at room temperature is 7? Pure water/neutral solutions are neutral because [H=] = [OH-] Using Kw = [H+][OH-] -> Kw = [H+]^2 so [H+] = (Kw)^1/2 At 25 degrees C, [H+] = (1 x 10^(-14))^1/2 = 1 x 10^7 so pH = 7
Calculate the pH of water at 50 degrees C given that Kw=5.476 x 10^(-14) [H+] = (5.476 x 10^(-14))^1/2 =2.34 x 10^7 pH = -log(2.34 x 10^7) =6.6 Still neutral as [H+] = [OH-]
Use Le Chatelier's principle to predict the change in pH when the temperature is increased The dissociation of water is endothermic so increasing the temperature would push the equilibrium to the right, giving a bigger concentration of H+ ions and thus a lower pH
How do we find the pH of a STRONG BASE, given the concentration of OH- ions? First work out [H+] by rearranging Kw = [H+][OH-] Then, use pH = -log [H+]
Weak acids only __________ when dissolved in water, giving an _________ mixture only slightly dissociate when dissolved in water, giving an equilibrium mixture
Give the more complex equilibrium equation showing what happens when a weak acid is dissolved in water HA + H2O(l) <----> H3O+(aq) + A-(aq)
Give the WEAK ACIDS dissociation expression
The larger the Ka the ______ ___ ____ stronger the acid
Rearrange pKa = -log Ka to make Ka the subjecy Ka = 10 ^ (-pKa)
What two assumptions are made to simplify the WEAK ACID dissociation equation to Ka = [H+]^2 / initial[HA (aq)] 1) [H=] = [OH-] because they dissociate according to a 1:1 ratio 2) As the amount of dissociation is small, we assume the initial conc. of the undissolved acid is unchanged
What is the pH of a solution of 0.01M ethanoic acid (Ka 1.7 x 10^-5)? Ka = [H+]^2 / [CH3CH2CO2H] [H+] = (1.7 x 10^(-5) x 0.01)^1/2 = 4.12 x 10^-4 pH = - log [H+] pH = - log (4.12 x 10^-4) pH = 3.38
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