# 0.4 Antacids  (Page 4/4)

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A primary standard is usually a solid reactant that:

(1) is available in a high purity form

(2) does not change chemically when stored or exposed to air

(3) has a high formula weight to minimize errors in weighing

(4) is soluble in the solvent being used.

Sodium carbonate, ${\text{Na}}_{2}{\text{CO}}_{3}$ , is commonly used as a primary standard base for standardizing acids, while potassium acid phthalate (KHP), ${\text{KHC}}_{8}{H}_{4}{O}_{4}$ , and oxalic acid dihydrate, ${H}_{2}{C}_{2}{O}_{4}\cdot {2H}_{2}O$ , are primary standard acids used for standardizing bases.

${\text{Na}}_{2}{\text{CO}}_{3}\left(s\right)+2\text{HCl}\to {H}_{2}O+{\text{CO}}_{2}+2\text{NaCl}$ (4)

${\text{KHC}}_{8}{H}_{4}{O}_{4}\left(s\right)+\text{NaOH}\to {\text{KNaC}}_{8}{H}_{4}{O}_{4}+{H}_{2}O$ (5)

$\left(\text{COOH}{\right)}_{2}\cdot {2H}_{2}O\left(s\right)+2\text{NaOH}\to {\text{Na}}_{2}\left(\text{COO}{\right)}_{2}+{4H}_{2}O$ (6)

The dry solid is carefully weighed on an analytical balance and then diluted in a volumetric flask to give a known molarity. The molarity of an acid or base is determined by titrating a measured volume of the acid or base with the primary standard. Then these acid or base solutions can in turn be used to determine the molarity of another acid or base (this is the case with the standardized solutions of NaOH and HCl used in this experiment).

In the example below a known amount of KHP will be titrated with a solution of NaOH to determine the NaOH solution concentration.

EXAMPLE: If 0.8168 g of KHP requires 39.35 mL of NaOH to reach the endpoint in a titration, what is the molarity of the NaOH? (1 mol KHP = 204.2 g)

$0\text{.}\text{8168}\text{gKHP}×\left(\frac{1\text{molKHP}}{\text{204}\text{.}2\text{gKHP}}\right)\left(\frac{1\text{molNaOH}}{1\text{molKHP}}\right)\left(\frac{1}{\text{39}\text{.}\text{35}\text{mLNaOH}}\right)\left(\frac{\text{1000}\text{mL}}{1L}\right)=0\text{.}\text{1016}\text{mol}/L=0\text{.}\text{1016}M$

## Notes on titration method:

1. Clean the buret until it drains smoothly.

2. Make sure the stopcock doesn't leak. The level should hold for ~3 minutes.

3. Remove air bubbles from buret tip before beginning. Rapid spurts usually work.

4. Rinse the buret two or three times with 5 mL portions of titrant (the solution you will use to titrate). Hold it horizontal and rotate to rinse.

5. Don't forget to record the initial reading. It does not need to be 0.00 mL.

6. Use a reading card to find the bottom of the meniscus.

7. Always read and record buret volumes to 0.01 mL.

8. Remember that the buret scale reads down, not up.

9. Put white paper under the flask to see color change easily.

10. Swirl the flask with one hand while turning the stopcock with the other or use stir bar and stir plate.

11. Add titrant slowly near the endpoint. Color that dissipates can be seen when getting close to endpoint.

12. Drops can be "split" by quickly turning the stopcock through the open position.

13. Near the endpoint use deionized water from wash bottle to rinse the flask walls from any splashed drops (adding water does not affect number of moles you have).

14. Don't drain buret below 25 mL (You won’t be able to determine final volume accurately). If necessary, read the buret level, refill, read the level, and continue.

## Calculation clarification

Amount neutralized by antacid 25ml of HCLThe amount of HCl that was neutralized by the fragment of the whole tablet, was only neutralized in part, this is where the titration comes in, the solution was still acidic, even after the dissolved antacid fragment, so the volume of NaOH used to finally cause the change in color is equal to the excess of HCl that was in the solution of HCl + water + antacid. So the number of moles of NaOH that was used is equal to the number of moles of HCl that was neutralized by the NaOH. So we DO use the formula in the procedure, so in the report form, where it had asked for the number of moles of HCl neutralized by the antacid:

${n}_{\text{HC}\text{ln}\text{eutralizedbyAntacid}}={n}_{\text{HCladded}}-{n}_{\text{HClexcess}}$

So then, this would be the $2\text{.}5×{\text{10}}^{-3}$ mol HCl–the excess which is calculated by converting the volume of NaOH used into moles, this number of moles then equals the number of moles oh HCl.

Volume NaOH titratedvolume excess HCl = volume titration of NaOH

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