7009431
Description
Flashcards by Anushka John, updated more than 1 year ago
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Created by Anushka John
about 8 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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