Question

If I dissolve a Tums tablet $(1000mg{\text{ }}CaC{O_3})$ in $25mL$ of water, how do I calculate the pH of the resultant suspension? When done experimentally, it was between $9$ and $10$.

Answer 1

We should understand that Tums tablet is a medicine and an antacid. It is used to treat stomach-related symptoms such as heartburn, upset stomach and indigestion. It works by lowering the amount of acid in the stomach thus curing the symptom.

Complete step by step answer:
Tums tablet contains calcium carbonate. We know that calcium carbonate is use as an antacid.
Calcium carbonate reacts with water, but it is considered to be insoluble in water as it is a salt.
We know that a saturated solution of calcium carbonate will hold $0.013g$ of dissociated salt in every $1L$ of water under standard temperature.
Thus, the solution will hold
$= Volum{e_{solution}} \times \dfrac{{mas{s_{dissociatedsalt}}}}{{{{10}^3}mL{\text{water}}}}$
$= 25m{L_{water}} \times \dfrac{{0.013{g_{calciumcarbonate}}}}{{{{10}^3}m{L_{water}}}}$
$= 0.000325gCaC{O_3}$
We will convert it into moles using molar mass concept.
$n = 0.000325g \times \dfrac{{1mol}}{{100.09g}}$
On simplification we get,
$n = 3.25 \times {10^{ - 6}}mol$
The calcium carbonate will dissociate in water to produce equal number of cations and anions.
$CaC{O_3} \rightleftharpoons C{a^{2 + }} + CO_3^{2 - }$
Thus, it implies that the saturated solution of calcium carbonate will contain $3.25 \times {10^{ - 6}}mol$ of carbonate anions in $25mL$ of water.
Now, the molarity of carbonate anions can be calculated as,
$M = \dfrac{{3.25 \times {{10}^{ - 6}}}}{{25 \times {{10}^{ - 3}}}}$
$M = 1.30 \times {10^{ - 4}}M$
Since, carbonate anion is a weak base, it will react with water to form bicarbonate anions and hydroxide anions.
$CO_3^{2 - } + {H_2}O \rightleftarrows HCO_3^ - + O{H^ - }$
We know that the acid dissociation constant of bicarbonate anion is,
${K_a}_{HCO_3^ - } = 4.8 \times {10^{ - 11}}$
Thus, the base dissociation constant of carbonate anion will be,
${K_b}_{CO_3^{2 - }} = \dfrac{{{{10}^{ - 14}}}}{{4.8 \times {{10}^{ - 11}}}}$
On simplification we get,
${K_b}_{CO_3^{2 - }} = 2.08 \times {10^{ - 4}}$
By definition of base dissociation constant, it can be expressed as,
${K_b}_{CO_3^{2 - }} = \dfrac{{[HCO_3^ - ].[O{H^ - }]}}{{[CO_3^{2 - }]}}$
Let us take $xM$ as the equilibrium concentration of hydroxide anions.
Thus, at equilibrium the solution will contain,
$[HCO_3^ - ] = [O{H^ - }] = xM$
$[CO_3^{2 - }] = (1.30 \times {10^{ - 4}} - x)M$
This is so, because in the reaction to produce $xM$ of hydroxide anions and bicarbonate anions, it should consume $xM$ of carbonate anions.
If we put the above two values in the equation of base dissociation constant, we get,
${K_b}_{CO_3^{2 - }} = \dfrac{{x.x}}{{1.30 \times {{10}^{ - 4}} - x}}$
$2.08 \times {10^{ - 4}} = \dfrac{{{x^2}}}{{1.30 \times {{10}^{ - 4}} - x}}$
Rearranging the equation to polynomial form,
${x^2} + 2.08 \times {10^{ - 4}}x - 2.08 \times 1.30 \times {10^{ - 8}} = 0$
Here, we will get two solutions. But we will neglect the negative one.
Thus, $x = 9.06 \times {10^{ - 5}}$
At equilibrium, the solution has,
$[O{H^ - }] = 9.06 \times {10^{ - 5}}M$
We will now calculate the pH of solution as,
$pH = 14 - [ - \log ([O{H^ - }])]$
Now we can substitute the known values we get,
$pH = 14 + \log (9.06 \times {10^{ - 5}})$
On simplification we get,
$pH = 9.96$
Thus, the given question states correct =. As practically the pH lies between 9 to 10. By calculation also, the pH lies between 9 to 10, $9.96$ to be exact.

Note:
We know that in physical chemistry, base dissociation constant is used to measure the basicity of a base in a solution. Or in simple words to measure the base strength or concentration of base in a given solution. The base dissociation constant is related to acid dissociation constant. If we sum up both together it will sum up to 14.