Acids + bases (Y13) Flashcards
remember to include the charges of ions which gain or loose a proton when writing acid-base equations
conjugate acid vs conjugate base
conjugate acid = accepts protons
conjugate base = lost a proton
conjugate acid of H2PO4-
H3PO4
which cannot function as a Bronsted-Lowry acid
CH3COO-
acid dissociation
1) remove the proton and write it as H+
2) Remember the charge on the conjugate base
Give an equation to show each stage in the dissociation of sulfuric acid in aqueous solution
H2SO4 (aq) –> H2SO4- (aq) + H+ (aq)
HSO4- (aq) –> SO42- + H+ (aq)
weak acid
barely dissociates
strong acid
fully dissociates
calculate the pH of the solution formed when 50cm3 of water is added to 100cm3 of 0.270mol dm-3 HCl
100 + 50 = 150
0.270 x (100 / 150) = 0.180
-log(0.18) = 0.74
calculate the concentration of a solution of H2SO4 with a pH of 0.84
this divide by 2
acid =
proton donor
base =
proton acceptor
acid-base equlibria
Acid-base equilibria involves the transfer of protons (H+ ions)
strong acids and bases
-strong acid = determined by the amount of H+ ions e.g H2SO4
-base = determined by amount of hydroxide (OH-) ions
-base + acid = salt + water (neutralisation reaction)
neutralisation ionic equation
H+ + OH –> H2O (neutralisation ionic equation)
Strong acids completely dissociate into ions in solution
equations pH
pH = –log10[H+] -> always give pH values to 2 decimal points
equation H+ ion
[H+] = 10^-pH
higher conc of H+ ions =
Higher concentration of H+ ions = lower pH (more acidic)
Bronsted-Lowry theory of acids
-Acid = substance that can donate a proton (H+)
-base = substance that can accept a proton
e.g HNO3 + NH3 –> NH4+ + NO3- (HNO3 donates a proton to NH3 so is the acid)
H+ ion dissociates from the acid
A H+ ion can only be accepted if a lone pair is present
There is 25cm3 of 0.1mol dm-3 of NaOH. This is neutralized by 32cm3 of H2SO4. Calculate the pH of the acid:
-moles of NaOH
-H+ + OH- –> H2O (1 mol in each so don’t multiply by 2)
-moles of H+ ions
-concentration of H+ ion
-pH = –log10[H+]
When calculating pH of an acid multiply by the no. Of moles e.g in H2SO4 but if in solution or dilute whatever then don’
If gives 2 vol e.g 250 and 50 in dilute solution add them –> 50/350 = volume
monotropic strong acid
The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid.
-For HCl and HNO3 the [H+ (aq)] will be the same as the original concentration of the acid.
excess acid
Work out new concentration of excess H+ ions [H+ ] = moles excess H+ total volume (dm3 ) pH = – log [H+ ] —> Total volume = vol of acid + vol of base added
35cm3 of 0.5 mol dm-3 H2SO4 is reacted with 30cm3 of 0.55 mol dm-3 NaOH. Calculate the pH of the resulting mixture.
Moles H2SO4 = conc x vol = 0.5 x 0.035 = 0.0175mol
Moles H+ = 0.0175 x2 = 0.035
Moles NaOH = mol OH- = conc x vol = 0.55 x 0.030 = 0.0165
H+ + OH- –> H2O Moles of H+ in excess = 0.035 -0.0165 = 0.0185 [H+ ] = moles excess H+ total volume (dm3 ) = 0.0185/ 0.065 = 0.28 mol dm-3
pH = – log [H+ ] = -log 0.28 = 0.55
45cm3 of 1.0 mol dm-3 HCl is reacted with 30cm3 of 0.65 mol dm-3 NaOH. Calculate the pH of the resulting mixture.
Moles HCl =mol H+ = conc x vol = 1 x 0.045 = 0.045mol
Moles NaOH =mol OH- = conc x vol = 0.65 x 0.030 = 0.0195
H+ + OH- –> H2O
Moles of H+ in excess = 0.045-0.0195 = 0.0255 [H+ ] = moles excess H+ total volume (dm3 ) = 0.0255/ 0.075 = 0.34 mol dm-3 pH = – log [H+ ] = -log 0.34 = 0.47
conc = mol2dm-6
square root concentration
explain why water is neutral at 50 degrees
[H+] = [OH-]
dissociation gives 1 H+ and 1 OH-
weak acid
doesnt fully dissociate
Explain why an aqueous solution containing propanoic acid and its sodium salt constitutes a buffer system able to minimise the effect of added hydrogen ions.
propanoic acid is a weak acid
Calculate the pH at 25 °C of the solution that results from mixing 19.0 cm3 of 2.00 M HCl with 16.0 cm3 of 2.50 M NaOH.
1) 0.019 x 2 = 0.038
2) 0.016 x 2.50 = 0.040
3) 19 + 16 = 35cm3 (total volume)
4) 0.040 - 0.038 = 2x10^-3
5) find concentration and put in equation
why is pure water always neutral
concentration of H+ ions is equal to the concentration of OH-
another way of displaying Kw
Kw = [H+] ^2
Kw at room temperature
1 x 10^-14
is the dissociation of water endothermic or exothermic
endothermic
what happens if we decrease the temperature of Kw
shifts left
favour backwards exothermic reaction
pH increases
Kw decreases
increase temperature of Kw=
becomes acidic (decrease pH)
if Ca(OH)2 or anything similar
multiply conc by 2 due to 2 moles
strong acids / monotropic basic
HCl
HNO3
weak acids / monotropic basic
carboxylic acids (COOH)
strong acid / diprotic basic
H2SO4
strong bases / monoprotic basic
NaOh
KOH
strong bases / diprotic basic
Ba (OH)2
weak bases / monoprotic basic
NH3
explain why H2O is not shown in the Kw expression
H2O is liquid = constant = not included (same for solids)
which statement about pH is correct
at temperatures above 298K the pH of pure water is less than 7
what is the concentration of NaOH that has a pH of 14.30
1) 10 ^ -14.30 = 5.011 x 10^-15
2) Kw = 1.00 x 10^-14
3) Kw / H+ = 2.00
why is pure water at 10 degrees not alkaline
[H+] = [OH-]
true or false -> water can act as an acid or a base
true
calculate H+ using Kw and OH-
Kw / [OH-]
what is meant by the term strong when describing an acid
the ability to fully dissociate and form H+ ions in water
water
water is slightly dissociated –> slightly ionised as it breaks apart but then forms again with other molecules. (polar, covalent molecule)
H2O reversible reaction
H2O ⇌ H+ + OH- ΔH = endothermic
kc [H2O]
Kc [H2O] = [H+] [OH-] –> the concentration of H2O is much greater than the concentration of H+ or OH- as water only slightly dissociates so we say it is approximately constant. (minor ionisation)
Kw calculation
Kc [H2O] = constant = Kw
Kw = [H+] [OH-]
temperature on Kw
-Only temperature affects Kw value –> forward reaction is endothermic
-Increasing the temperature of Kw will favour the forward endothermic reaction to oppose the increase in temperature. The yield of H+ and OH- will increase so Kw increases.
increasing temperature on Kw
Increase in temperature = increase in ions = decrease in pH but the solution is still neutral as [H+] = [OH-] –> increase in temp also increases ionisation energy so there is more energy to break bonds
concentration of 1.0 x10^-14 find pH
-square root concentration = 1.0 x 10^-7
=put in log equation
= pH of 7
Rearranged equation to find [H+] = Kw / [OH-]
Find pH of 0.200 moldm-3 of NaOH:
-[OH-] = 0.200
-Kw = 1.0 x 10^-14 / 0.200 = 5 x10^-14
–log (5 x 10^-14) = 13.30
Find the concentration of Ba(OH2) with a pH of 13.30:
-10^-(13.30) = 5.01 x 10^-4 = H+
-Kw / H+ = 0.200
-0.200 / 2 = 0.100
-you have to divide by 2 because there are 2 OH molecules
Find the pH of 0.15mol dm-3 KOH:
-log (0.15) = 0.8239….
-14 – 0.823… = 13.18
-you have to subtract from 14 as total pH is 14
Kw value =
1 x 10^-14 mol2dm-6
calculate pH of H2SO4
multiply concentration by 2 before putting in log
square root value of Kw
-only when the concentration of ions in a neutral solution e,g pure water
pH of a diluted solution
vol = 0.3
HCl = 0.8
pH = 10.0
added 0.3dm3 of water
1) Find new conc of H+ ions
2) pH of diluted solution
0.3 x 0.8 = 0.24 H+ ions
New vol = 0.3 + 0.3 = 0.6
new [H+] = 0.24 / 0.6 = 0.4
dilute HCl in water
lower conc of H+ = increase in pH
conc of H+ =
conc of acid
pH change with solid bases
Find pH, concentration of H+, volume, moles of acid and moles of H+ ions for original acid
1) volume remains the same
2) Moles = subtract from reacting base
3) moles H+ = moles of resulting solution
4) moles of H+ x unchanged volume = concentration of H+
5) find pH using log
add solid base to solution of an acid
pH will increase
volume of solution will remain the same
add aqueous base to solution of an acid
volume increases
H+ ions decreases
Ezra has 0.25dm3 of HCl with concentration of 0.2moldm-3.
He reacts 0.04 moles of NaOH dissolved in 0.12dm3 of water.
How many H+ ions left when the reaction has finished
And what is the volume of the resulting solution
Part 1:
0.25 x 0.2 = 0.05
0.05 - 0.04 = 0.01
Part 2:
-0.37
calculating new pH of adding aqueous base
1) find change in volume
2) find change in {H+]
find new pH of solution after dilution
30ml HNO3
0.2moldm-3
20ml H2O
1) 0.2 x (30/1000) = 0.006
2) add the 2 volumes = 50ml
Conc of H+ ions = moles of H+ ions / volume of solution
0.006 / (50 / 1000) = 0.12
put in log
pH of solution after the reaction
50ml HBr
0.30moldm-3
0.120 moles KOH
1) moles = 0.015
2) moles of HBr - moles of KOH = 0.003
3) 0.003 / volume of HBr
4)0.06
5) put in log
doesnt rlly matter if you divide or multiply by 2
easier to multiply though
steps to calculating pH of mixed solutions
1) mol of H+
2) mol of OH-
3) subtract excess (big number) from limiting to get complete moles
4) excess value / total volume
5) put in log equation to get pH for acid in excess
6) if finding pH of base in excess subtract log answer from 14 or put in Kw equation
Calculate the pH of a solution formed when 50cm3 of 0.100mol dm-3 H2SO4 is added to 25cm3 of 0.150mol dm-3
-mol of H+ = 2 x 50/1000 x 0.100 = 0.01
-mol of OH- = 0.025 x 0.150 = 0.00375
-excess = H+ as it is larger so 0.01 - 0.00375 = 0.00625
-0.00625 / (50+25/1000) = 0.0833
-log (0.0833) = 1.079
pH of 25cm3 of 0.25moldm-3 H2SO4. 100cm3 of 0.2moldm-3 NaOH
H+ = 0.0125 moles
-OH- = 0.0200 moles
-OH = in excess (0.0200 - 0.0125) = 0.0075
-0.0075 / 125/1000 = 0.06
–log = 1.22
-14 – 1.22 = 12.78
when calculating change in pH
-Do normal steps to find pH of excess reagent
-Find original pH of acid and subtract values
-When finding pH change always subtract new value of pH from original value of pH acid
10.35cm3 of 0.100moldm-3 HCl added to 25cm3 of 0.150moldm-3 Ba(OH)2. Calcuate pH of solution at 30 degrees with Kw of 1.47 x 10-14.
1) H+ moles = 1.035 x 10^-3
2) OH- moles = 7.5x10^-3
3) 7.5 - 1.035 = 6.465 x 10^-3
4) 6.456 x10^-3 / (10.35 + 25) / 1000 = 0.1828854314…
5) (1.47 x 10^-14) / (0.182….) = 8.038 x 10^-14
6) -log (ANS) = 13.09
Calculate the pH of the solution formed after the addition of 35cm3 of 0.150moldm-3 NaOH to the original 25cm3 of monotropic acid:
1) moles NaOH = 5.25 x 10^-3
2) moles of montropic acid (assuming conc is same as base) = 3.75 x 10^-3
3) difference = 1.5 x 10^-3
4) assume OH- is in excess
5) (3.75 x 10^-3) / (35+25) / 1000 = 0.0625 000 = 0.0625
6) Kw / 0.0625 = 1.6 x 10^-3
7) -log (1.6 x 10^-3) = 12.80
propanoic acid =
C3H6O2
1 carbon counts as part of the carboxyl group
weak acids
-weak acids and weak bases dissociate slightly in aqueous solutions
Ka
Ka = dissociation constant of a weak acid
equations for a weak acid
PKa = -log (Ka)
Ka = [H+] [A-] / [HA]
Ka = 10^-pKa
calculate Ka
10^-pka