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Get Started Now- Intro Lesson: a18:10
- Intro Lesson: b8:08
- Intro Lesson: c4:54
- Intro Lesson: d9:25
- Lesson: 1a3:34
- Lesson: 1b4:34
- Lesson: 27:49

In this lesson, we will learn:

- Why titrations are performed and their purpose as an experiment.
- How to use reaction stoichiometry to find moles and unknown concentrations from titrations.
- How to use titrations and mole calculations to find percentage purity of a sample.

__A__. We can use an acid or base sample with a known concentration (let’s call the ‘known’ chemical A) in a chemical reaction with a sample of a base or acid ofis a lab experiment used to find the unknown concentration of an acid or base*titration*concentration that we’re investigating (let’s call the ‘unknown’ chemical B).*unknown*

Chemicals A and B will react in proportional amounts (moles) to each other, so the amounts of A we used, in the end, tells us the amounts of B that’s present.__Titration experiments work because volume AND concentration of chemical A are known, so we can calculate moles of A, which is related to moles of B (in their reaction), which is used with its volume to calculate A’s unknown concentration!__Finding this works in a sequence as shown in the table below:

- The practical experiment goes like this:
- Known chemical A is gradually added to a precise volume of unknown B until the
*stoichiometric point*is reached. The stoichiometric point is when the ratio of moles of A to B is the same as in the balanced chemical equation. This point is clear from a large change in pH (and color change with indicator added) because*neutralization*occurs.__The molar ratio in the reaction equation (the stoichiometry) and the volume of A that was needed to neutralize B is then used to find the number of moles of B__. Now that moles and volume of ‘unknown’ B are known, the__concentration of the unknown chemical B sample can be calculated__.

- Known chemical A is gradually added to a precise volume of unknown B until the
- Titrations are excellent for finding acid/base concentrations because:
- Acids and bases are complementary in their chemistry so there are lots of ideal ‘partner’ chemicals for them.
- It is easy to measure pH and pH change of a mixture, and the pH can be used to find the concentration and moles of H
_{3}O^{+}or OH^{-}in that same mixture. This is perfect for chemicals that produce these ions, like acids and bases!

- The most important factor in a successful titration is
__precision__. The titration needs to stop precisely at the stoichiometric point (AKA*equivalence point*). To make sure this is accurate and precise, a few steps are taken in the practical method:- A chemical called an
__indicator is added to the mixture__. This chemical__indicates a change in pH by a clear color change, and the equivalence point is exactly when the titration mixture changes color permanently__. There are many different indicators for different pH ranges and combinations of weak/strong acids reacting with weak/strong bases. They are very sensitive to pH change – the addition of one extra drop of acid or base might cause a permanent color change. - A
__trial run__is performed on the very first titration experiment. This is a ‘rough run’ to establish roughly the volume of chemical A needed to reach the equivalence point. This is done to save time and still allow chemists to be precise in the approximate range where the color change – and equivalence point – is found. - The final amounts of the known chemical A are added dropwise to get as precise a reading as possible. A single titration run with a volume reading obtained is called a ‘titer’.
- The precise runs – titers – are repeated until two readings fall within 0.05cm
^{3}of each other. An average titer is then taken from these two readings.

- A chemical called an
- Once an average titer has been obtained, this is taken to be the volume where there is an equal number of moles of both acid and base.
__Remember that your titer volumes will be in milliliters or cm__. You must^{3}but concentration is measured in moles per LITER, or dm^{3}__convert__the volume to liters! Dividing milliliters by 1000 will give you a value in liters.

The number of moles of chemical B can be calculated because its concentration and volume are both known:$concentration \; (M) = \frac{moles\; (mol)}{Volume \; (L)}$

Rearrange to give:$moles \; (mol) = concentration \;(M)\; * \; Volume \; (L)$

With this moles of B value obtained, recall the reaction stoichiometry:aA + bB $\;$→$\;$ cC + dD

The molar ratio is important for finding moles of B.__The equivalence point occurs when b moles of A is equal to a moles of B__. Use this to find the number of moles of unknown chemical B in the titration mixture at equivalence point. - Once the number of moles of chemical B is found, you have moles and volume:
$concentration \; (M) = \frac{moles\; (mol)}{Volume \; (L)}$

Now the unknown concentration can be calculated! - Titrations involving
__partial neutralization__can also be performed; this is where polyprotic acids (acids with more than one proton to donate) donate a proton one by one, so no one step completely neutralizes the acid. Take the example with phosphoric acid, H_{3}PO_{4}:Reaction 1: H _{3}PO_{4}+ OH^{-}$\;$→$\;$ H_{2}PO_{4}^{-}+ H_{2}OReaction 2: H _{3}PO_{4}+ 2OH^{-}$\;$→$\;$ HPO_{4}^{2-}+ 2H_{2}OReaction 3: H _{3}PO_{4}+ 3OH^{-}$\;$→$\;$ PO_{4}^{3-}+ 3H_{2}O

If you are asked to do a calculation with__partial neutralization__, you__might need to find out the acid:base molar ratio__in the reaction. For this, you’ll be given the volume and concentration of both acid and base so that moles of both can be found – then divide the larger number of moles by the smaller to find the ratio between them! To find the moles, the rearranged equation for concentration can still be used:$moles \; (mol) = concentration \;(M)\; * \; Volume \; (L)$

This is sometimes written as:n = C * V $\quad$ Where $\,$ n = moles, C = concentration (M), V = volume (L *or*dm^{3}) __Titration results can be used to find the__of a sample as well. This is done in the following method:*percentage purity*- Find the number of moles of the acid/base you added to the solvent to create the solution. For example, a 1.13g sample of NaOH was used. Find the number of moles of this by dividing the mass used by molar mass:
$moles = \frac{mass \; (g)}{molar \; mass\; (gmol^{-1})}$ $moles = \frac{1.13}{40} = 0.028 mol$

Find the__expected concentration__by dividing the volume of the solution made by this number of moles.

For example, the NaOH used was diluted to 250 mL (0.25 L):$concentration\; (M) = \frac{moles \; (mol)}{volume\; (L)}$ $concentration\; (M) =\frac{0.028\; mol}{0.25 (L)} = 0.112\; M$ __Record your expected concentration__. - Perform the titration and find the concentration of your sample substance using your titer results with the method above. This is your
__actual concentration__– if you have impurities in your sample, they won’t react but they still took up mass in your sample, so this actual concentration will be lower than the expected concentration.

For example, a 25.00 mL sample of the NaOH solution was titrated with 27.90 mL (or 0.0279 L) of 0.097 M HCl:$moles \; (mol) \; = 0.097M * 0.0279L = 0.0027 mol$

HCl and NaOH react in a 1:1 ratio, so 0.0027 mol HCl = mol NaOH:$concentration \; (M) = \frac{0.0027\; mol}{0.025\; L} = 0.108M$ - Use these two values to calculate percentage purity of your sample with the equation:
$\% \; purity = \frac{actual \; concentration}{expected \; concentration} * 100= \frac{1.08\;M}{1.12\;M} * 100 = 96.4 \%$

- Find the number of moles of the acid/base you added to the solvent to create the solution. For example, a 1.13g sample of NaOH was used. Find the number of moles of this by dividing the mass used by molar mass:

- Introduction
__What is titration?__a)How titration worksb)Finding unknown concentration: titration calculations.c)Titrations with partial neutralization.d)Percentage purity. - 1.
**Determine the unknown concentration from acid-base titration results:**

A solution of KOH of unknown concentration was titrated by 0.3 M hydrochloric acid. The results below show the volume of HCl required to titrate 25cm^{3}of the KOH solution:

Run:

Titre volume (cm

^{3})1

18.60

2

18.35

3

18.40

a)- Which two of these readings are concordant and should be used to find the average titre?
- Calculate the average titre for this experiment.

b)Use your average titre to find the concentration of the potassium hydroxide (KOH) solution. - 2.
**Use an acid-base titration to calculate percentage purity of a sample.**

1.17g of sodium hydroxide pellets were used to make a 250mL solution of NaOH. 25.00 mL of this solution was titrated by 0.096 M HNO_{3}, with an average titre of 27.75 mL.

Use this information to find the percentage purity of the sodium hydroxide sample.

2.

Acid-Base Theory

2.1

Introduction to acid-base theory

2.2

Conjugate acids and bases

2.3

Strong and weak acids and bases

2.4

pH and pOH

2.5

Acid dissociation constant

2.6

Relative strength of acids and bases

2.7

Ionic product of water: K_{w}

2.8

K_{a} and K_{b} calculations

2.9

Acid-base titration

2.10

pH indicators

2.11

Titration curves

2.12

Buffer solutions