Download presentation

Presentation is loading. Please wait.

Published byTaylor Milledge Modified over 3 years ago

1
Acid and Base Titration calculations 7-5

2
Titration calculations Chemists use the fact that the pH breaks sharply when an acid is neutralized by a base to accurately determine the concentration of either the acid or base solution. At this point there is stoichiometrically equal amounts of acid and base.

3
Molarity = moles Liter Moles of a substance =Concentration (M) x Volume (L)

4
Balance Base (Bb) x Moles of Acid = Moles of Base x Balance Acid (Ba) Bb x Ca x Va = Ba x Cb x Vb

5
equivalence point This point can be determined within a few drops of base solution added, because the pH changes sharply near this point. is when there is an equal amount of moles of acid and base

6
Chemists use an indicator to signal the pH change. An indicator is a substance that changes color with certain changes in pH. The experimenter picks a indicator that changes color in the pH range that corresponds to the break in the titration curve. This will vary depending on what combinations of acids and bases are titrated. When the indicator changes color the titration is said to have reached it's endpoint. If the titration is done properly the endpoint is very close to the equivalence point.

7
Steps for the titrations Step 1: balance the neutralization equation. determine balance of acid and base. Step 2: determine the given information. Step 3: determine what information is required Step 4: solve using the equation below Bb x Ca x Va = Ba x Cb x Vb

8
example: Calculate the concentration of a nitric acid solution (HNO 3 ) if a 20. ml sample of the acid required an average volume of 55 ml of a 0.047 mol/l solution of Ba(OH) 2 to reach the endpoint of the titration.

9
Step 1: 2 HNO 3 + Ba(OH) 2 -----> Ba(NO 3 ) 2 + 2 H 2 O Balance Base = 1 Balance Acid = 2 Look at the coefficients to decide these numbers acid base

10
Step 2: Volume Acid = 20 ml Volume Base (average) = 55 ml Concentration of Base = 0.047 mol/l

11
Step 3: Concentration of the acid

12
Step 4: Solve using the equation. Bb x Ca x Va = Ba x Cb x Vb (1)( Ca)(20 ml) = (2)(0.047 mol/l)(55 ml) Ca = 0.2585 mol/l Ca = 0.26 M

13
example: Calculate the concentration of a sulfuric acid solution (H 2 SO 4 ) if a 40. ml sample of the acid required an average volume of 125 ml of a 0.096 M solution of NaOH to reach the endpoint of the titration.

14
H 2 SO 4 + 2 NaOH → 2 H 2 O + Na 2 SO 4 Bb x Ca x Va = Ba x Cb x Vb (2)(Ca)(40 ml) = (1)(0.096 mol/l)(125 ml) Ca = 0.15 mol/l acid base

15
Solve using the equation. Bb x Ca x Va = Ba x Cb x Vb (1)( Ca)(20 ml) = (2)(0.047 mol/l)(55 ml) Ca = 0.2585 mol/l Ca = 0.26 M

Similar presentations

OK

THE MATHEMATICS IN A TITRATION CURVE (WITH A LITTLE BASE 10 AND LOGARITHM ARITHMATIC ADDED)

THE MATHEMATICS IN A TITRATION CURVE (WITH A LITTLE BASE 10 AND LOGARITHM ARITHMATIC ADDED)

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Free download ppt on conservation of plants and animals Download ppt on water a precious resource Ppt on network switching methods Download ppt on wildlife sanctuaries in india Maths ppt on areas related to circles Ppt on simple carburetor operation Ppt on forward rate agreement notation Ppt on manas national park Ppt on network switching table Ppt on network switching concepts