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(4) is soluble in the solvent being used.

Sodium carbonate, Na 2 CO 3 size 12{ ital "Na" rSub { size 8{2} } ital "CO" rSub { size 8{3} } } {} , is commonly used as a primary standard base for standardizing acids, while potassium acid phthalate (KHP), KHC 8 H 4 O 4 size 12{ ital "KHC" rSub { size 8{8} } H rSub { size 8{4} } O rSub { size 8{4} } } {} , and oxalic acid dihydrate, H 2 C 2 O 4 2H 2 O size 12{H rSub { size 8{2} } C rSub { size 8{2} } O rSub { size 8{4} } cdot 2H rSub { size 8{2} } O} {} , are primary standard acids used for standardizing bases.

Na 2 CO 3 ( s ) + 2 HCl H 2 O + CO 2 + 2 NaCl size 12{ ital "Na" rSub { size 8{2} } ital "CO" rSub { size 8{3} } \( s \) +2 ital "HCl" rightarrow H rSub { size 8{2} } O+ ital "CO" rSub { size 8{2} } +2 ital "NaCl"} {} (4)

KHC 8 H 4 O 4 ( s ) + NaOH KNaC 8 H 4 O 4 + H 2 O size 12{ ital "KHC" rSub { size 8{8} } H rSub { size 8{4} } O rSub { size 8{4} } \( s \) + ital "NaOH" rightarrow ital "KNaC" rSub { size 8{8} } H rSub { size 8{4} } O rSub { size 8{4} } +H rSub { size 8{2} } O} {} (5)

( COOH ) 2 2H 2 O ( s ) + 2 NaOH Na 2 ( COO ) 2 + 4H 2 O size 12{ \( ital "COOH" \) rSub { size 8{2} } cdot 2H rSub { size 8{2} } O \( s \) +2 ital "NaOH" rightarrow ital "Na" rSub { size 8{2} } \( ital "COO" \) rSub { size 8{2} } +4H rSub { size 8{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 . 8168 gKHP × 1 molKHP 204 . 2 gKHP 1 molNaOH 1 molKHP 1 39 . 35 mLNaOH 1000 mL 1L = 0 . 1016 mol / L = 0 . 1016 M size 12{0 "." "8168" ital "gKHP" times left ( { {1 ital "molKHP"} over {"204" "." 2 ital "gKHP"} } right ) left ( { {1 ital "molNaOH"} over {1 ital "molKHP"} } right ) left ( { {1} over {"39" "." "35" ital "mLNaOH"} } right ) left ( { {"1000" ital "mL"} over {1L} } right )=0 "." "1016" ital "mol"/L=0 "." "1016"M} {}

Notes on titration method:

1. Clean the burette until it drains smoothly.

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

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

4. Rinse the burette 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 burette volumes to 0.01 mL.

8. Remember that the burette 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 burette below 25 mL (You won’t be able to determine final volume accurately). If necessary, read the burette level, refill, read the level, and continue. 

Calculation clarification

Only part of the HCl is neutralized by the fragment of the antacid tablet. Since the solution is still acidic, NaOH is used to finally cause the change in color. The number of moles of NaOH is equal to the excess of HCl that was not neutralized by the antacid. We can use the formula in the procedure to fill in the part of the report form where it asks for the number of moles of HCl neutralized by the antacid:

n HCl neutralized by Antacid = n HCl added n HCl excess size 12{n rSub { size 8{ ital "HCl"` ital "neutralized"` ital "by"` ital "Antacid"} } =n rSub { size 8{ ital "HCl"` ital "added"} } - n rSub { size 8{ ital "HCl"` ital "excess"} } } {}

So then, this would be the 2 . 5 × 10 3 size 12{2 "." 5 times "10" rSup { size 8{ - 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 of HCl.

25ml of HCLVolume titrated by NaOHNaOHVolume titrated byantacid

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Source:  OpenStax, General chemistry lab spring. OpenStax CNX. Apr 03, 2009 Download for free at http://cnx.org/content/col10506/1.56
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