194 DIY Science: Illustrated Guide to Home Chemistry Experiments
SBSTITUTIU oNS ANd modIfICATIoNS- You may substitute foam cups or similar containers for the beakers.
- You may substitute a 100 mL graduated cylinder for the 100 mL volumetric flask and a 10 mL graduated cylinder for the
10 mL pipette, with some loss of accuracy. (Unless your pH meter is very accurate, instrumental error will be greater than
volumetric error anyway.) - You may substitute narrow-range pH testing paper for the pH meter, with some loss of accuracy. (Typical inexpensive pH
meters are accurate to 0.1 or 0.2 pH, while typical narrow-range pH testing paper is accurate to 0.4 or 0.5 pH.) - You may reduce the quantities of the acid and base solutions, depending on the size and type of probe used by your pH
meter, and use smaller containers. If your meter has a very thin probe (or if you are using pH paper), you can reduce the
quantities from 100 mL to 10 mL and use test tubes. - If you don’t have a 1 M bench solution of hydrochloric acid available, you can make it up by diluting 8.33 mL of 12 M (37%)
concentrated hydrochloric acid to 100 mL. If you have hardware-store muriatic acid that lists the contents as 31.45% HCl,
you can make up a 1 M solution by diluting 9.71 mL of acid to 100 mL. - If you don’t have a 1 M bench solution of sulfuric acid available, you can make it up by diluting 5.56 mL of 18 M (98%)
concentrated sulfuric acid to 100 mL. If you have hardware-store battery acid that lists the contents as 35% sulfuric acid,
you can make up a 1 M solution by diluting 15.57 mL of acid to 100 mL. - If you don’t have a 1 M bench solution of acetic acid available, you can make it up by diluting 5.75 mL of 17.4 M (99.8%)
concentrated (glacial) acetic acid to 100 mL. Alternatively, most distilled white vinegar contains 5% to 6% acetic acid,
which is close enough to 1 M to use for this experiment. - If you don’t have a 1 M bench solution of sodium hydroxide available, you can make it up by dissolving 4.00 g of sodium
hydroxide in water and making up the solution to 100 mL. - If you don’t have a 1 M bench solution of sodium carbonate available, you can make it up by dissolving 10.60 g of anhydrous
sodium carbonate in water and making up the solution to 100 mL. - I suggest boiling and cooling the distilled or deionized water before use to eliminate any dissolved carbon dioxide gas,
which forms carbonic acid and lowers the pH below the 7.00 pH of pure water. Alternatively, you can use distilled or
deionized water from a freshly opened container or one that has been kept tightly capped.
remains undissociated. Based on the previous calculation, we
know that the pH of a 1.0 M solution would be about 0 if the acid
had fully dissociated, so we know that the actual pH of the 1.0 M
acetic acid will have some value higher than 0.
If we assume for a moment that at equilibrium only 10% of the
acetic acid has dissociated, we can calculate the approximate pH
of the solution. If only 10% of the acid has dissociated in a 1.0 M
solution of acetic acid, the hydronium ion concentration is 0.1 M.
Filling in the formula, we get:
pH = –log 10 [0.1] = 1
But to estimate the pH of this solution accurately, we need
a better value than our guesstimate of 10% dissociation. We
get that value by looking up the equilibrium constant for the
dissociation reaction shown above. In the context of pH, the
equilibrium constant is referred to as the acidity constant,
acid dissociation constant, or acid ionization constant, and is
abbreviated Ka.
With the acid dissociation constant for acetic acid known to be
1.74·10–5, and ignoring the tiny contribution to [H 3 O+] made by
the water, we can calculate the pH of a 1.0 M solution of acetic
acid as follows:ka = 1.74·10–5 = ([H 3 o+]·[CH 3 Coo–]/[CH 3 CooH])An unknown amount of the acetic acid has dissociated, which
we’ll call x. That means that the concentration of the acetic
acid, or [CH 3 COOH] is (1.0 – x), while the concentrations of the
dissociated ions, [H 3 O+] and [CH 3 COO-], are both x. Filling in the
formula gives us:1.74 · 10–5 = ([x] · [x]/[1.0 – x])
or
1.74 · 10–5 = x^2 /(1.0 – x)
or
(1.74 · 10–5) · (1.0 – x) = x^2
or
1.74 · 10–5 – (1.74 · 10–5 · x) = x^2