pH 7 Add additional 2 drops of 0.500M NaOH, meas 8 Repeat until total of 12 drop
ID: 1005021 • Letter: P
Question
pH 7 Add additional 2 drops of 0.500M NaOH, meas 8 Repeat until total of 12 drops are added. Add 2 t6-8 for beaker 2, and then for beaker 3. Make sure to rinse the 9. Repeat 6-8 for be probe between beakers. Data Tables: Drops of PH Beaker 1 pH Beaker 2 pH Beaker 3 HCI added Acetic Acid+Ammonia + Acetate Ammonium Water 0 drops 67 g.25 6.87 2.46 95 4.24 26 9y .93. 1.80 12 Drops of PH Beaker 1 pH Beaker 2 pH Beaker 3 NaOH added Acetic AcidAmmonia+ Acetate .G5 Ammonium 0 drops Water 93 14.55 ·63 4 42 9.so opsExplanation / Answer
1] A buffer is a solution that can maintain a nearly constant pH if it is diluted, or if relatively small amounts of strong acids or bases are added.
Buffer solutions resist pH changes
From observing the given data,
Buffer 3rd is worse than both 1 and 2 buffers because it changes its pH upon the addition of small amounts of acids or bases.
Buffer 2 is a little bit bettter than buffer 1 because it changes pH very little .
Buffer 2 > Buffer 1 > Buffer 3 order of best ones.
2] when HCl is added
HCl + CH3COO- <==> Cl- + CH3COOH [acetate takes H+ from HCl and form acetic acid ]
when NaOH is added
CH3COOH + OH- <==> H2O + CH3COO- [OH- ion takes H+ from acid and from water ]
3]Ammonia / Ammonium Buffer
awhen HCl is added
NH3 + HCl ------> NH4+ [Forms ammonium ion by taking H+ from the acid ]
When NaOH is added
NaOH + NH4+ ------> NH3 + H2O [NaOH takes H+ from ammonium and forms water ]
4] Hydrogen chloride is a gas. When it dissolves in distilled water it becomes hydrochloric acid. Because it is a strong acid, it disassociated into H+ and Cl- ions. The solution process is very exothermic.If you add water to acid, you form an extremely concentrated solution of acid initially. So much heat is released that the solution may boil very violently, splashing concentrated acid out of the container!
NaOH simple ionizes into Na+ and OH- ions
5] Water is the big exception to being a buffer and we can understand this on the basis of La Chatlier
consider adding some .1M K+ OH- to water
H2O <------> OH- + H+ (rnx 1)
where [OH-]*[H+] = 10^-14
When we add some OH- to water you might think in terms of La Chatlier that the OH- will tie up the H+ in terms of the original acid (H2O) and produce more H+ to act as a buffer - but stare hard at rnx 1 you are adding OH- and by LaChaltlier it wants to make more H2O not less.
you dont have say a conjugate base different than OH- like C2H3O2-
Another way of saying this is when OH- is added to a buffer like the acetate mixture
because we have CH3COOh <----> H+ + CH3COO- when we tie up the H+ with OH- the system will make more H+ to resist the change by ionizing acetic acid.
Water cant do this trick because when you ionize water it makes more OH-
6] add some little amounts of acid and check pH
add some little amounts of base and check pH
if pH of the buffer doesnt change with both of them , then it is working .
7] Going beyond buffering capacity will cause an outlier of the graph as the bufferdoes not incorporate well when itOnce the buffering capacity is exceeded the rate ofpHchange quickly jumps. This occurs because the conjugate acid or base has been drainedthroughneutralizationprocess.This principle implies that a larger amount of conjugateacid or base will have a greater buffering capacity. In this experiment, buffer capacityhappens every time water reacts with the acidic and basic substance. When water reactswith HCl, the pH drops significantly from 7.46 to 3.05, which shows pH change of -4.41.When water reacts with NaOH, the pH significantly rises from 7.31 to 10.77, whichshows a positive pH change of 3.46. Apart from that, the buffer capacity also happenswhen ammounium buffer reacts with 12 drops of HCl, as the pH changes from 8.88 to8.25, which is .63 lower than the stable rate. This shows that the pH level dropsignificantly from the equivalent point as it is depleted from the neutralization occurrence.
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