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The pH scale The pH scale is used to measure how acidic or basic a solution is. The lower the pH, the more acidic the solution. The higher the pH, the more basic the solution.

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**pH – The power of Hydrogen**

The mathematical formula for pH is: pH = - log [H+] * * Remember H+ and H3O+ are both symbols for the hydronium ion. The symbol [ H+ ] stands for the concentration of hydrogen ion. The log of a number is the exponent that ten must be raised to give you the number: example: log (100) = log (102) = 2 log (1000) = log (103) = 3

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**Calculating pH from [H+]**

If the concentration of hydrogen ion is known, you can calculate the pH using the equation: pH = - log [H+] Calculate the pH for the following solution: [H+] = 1 x 10-3 M pH = - log (1x10-3) = -(-3) = 3 [H+] = 1 x 10-8 M pH = - log (1x10-8) = -(-8) = 8

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pH Practice Determine the pH of the following solutions and identify them as acidic, basic, or neutral. [H+] = 1 x 10-2 M pH = [H+] = 1 x M [H+] = 1 x M [H+] = 1 x 10-5 M pH = [H+] = 1 x 10-7 M 5 acidic 2 acidic 7 neutral 11 basic 13 basic

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pH indicates [H3O+] If the pH is know, the [H3O+] of the solution can be determined. Example: pH = 6 (acidic) [H+] = 1 x 10-6 M pH = 11 (basic) [H+] = 1 x M

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**Pure Water [H+] = [OH-] = 1 x 10-7 M**

Pure water has a pH of 7, which indicates a [H+] of 1 x 10-7 M. Pure water is neutral because it contains EQUAL amounts of acid and base ions. In pure water: [H+] = [OH-] = 1 x 10-7 M

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**Self-ionization of water**

Pure water undergoes a process called self-ionization in which a few molecules of water break up into ions according to the following equation: HOH + HOH H3O+ + OH- Since this reaction produces H+ * and OH- ions in equal amounts, the concentration of the two ions must be the same. * Remember H+ and H30+ are just different ways of abbreviating the hydronium ion.

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**An important relationship between hydronium and hydroxide ions**

Chemists have discovered that ANY time water is present, the following relationship will be true: [H+] x [OH-] = 1 x 10-14 We have seen that this relationship holds true with pure water, (1 x 10-7) x (1 x 10-7) = 1 x 10-14 But it also holds any time water is present, regardless of the pH of the solution.

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**Using the H+/OH- Relationship**

pH [H+] x [OH-] = 1 x 10-14 7 1x10-7 1x10-14 4 1x10-4 1x10-10 9 1x10-9 1x10-5

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**Now you try! pH [H+] x [OH-] = 1 x 10-14 8 1x10-8 1x10-? 1x10-14 5**

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Acid/Base Indicators Acid/base indicators are substances that have different colors at different pH’s. One of the most common indicators is litmus, which is red in acidic pH’s (less than 7) and blue in basic pH’s (greater than 7).

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Phenolphthalein Phenolphtalein is an important indicator that is used to determine when the acid in a neutralization reaction has been used up. When this happens, the color quickly changes from clear to a pinkish, magenta color. Before endpoint pH <8.2 At endpoint pH = 9 Overshoot pH >10

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**Using indicators to determine pH**

The chart above is a graphical representation of some of the indicators described in Table M.

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**Using indicators to determine pH**

Suppose we test a solution and find out that phenolphthalein turns pink in the solution, but indigo carmine is blue. What is the approximate pH of the solution?

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**Using indicators to determine pH**

In another case, we find a solution turns methyl orange yellow, but methyl red turns red. What is the pH of the solution? Between 4 & 5

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**Using indicators to determine pH**

Finally, we find a solution that turns bromothymol blue yellow, but methyl red is also yellow. What is the pH of the solution? Approximately 6

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Titration Titrations are procedures done to determine the amount of acid or base present in a solution.

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**Titrations 2 Sufficient base is added to the acid to neutralize it.**

At the endpoint, the color changes indicating that just enough base has been added to neutralize the acid

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Titration 3 Based on the volume of the acid, the concentration of the base, and the amount of base added, the concentration of the acid can be determined using the neutralization reaction.

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**Sample Titration Problem**

20 ml of vinegar (HC2H3O2) are titrated with .5 M NaOH. 25 mL of NaOH are required to reach the endpoint of the titration. What is the molarity of the acetic acid solution? As the base is added to the reaction, the NaOH reacts with the HCl according to the neutralization reaction: Acid + Base Salt + Water At the endpoint just enough base is added to neutralize the acid present, causing the indicator to change color.

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**Moles of NaOH = Molarity x volume (in Liters)**

Titration Problem 2 For this specific reaction: HC2H3O2 + NaOH NaC2H3O2 + HOH Using the volume of the NaOH and its molarity the number of moles of NaOH added can be calculated: Moles of NaOH = Molarity x volume (in Liters) = .25 M x .025 L = moles of NaOH

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**Molarity = moles / liters**

Titration Problem 3 However, the balanced chemical equation tells us that if moles of NaOH reacted, then moles of HC2H3O2 must have also reacted. The molarity of the acid can be calculated using the number of moles of acid neutralized by the base and the original volume of the acid placed in the Erlenmeyer flask. Molarity = moles / liters =.0625 moles/ .020 L = Molar

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**molarityacid x volumeacid = molaritybase x volumebase**

Titration Shortcut For titrations between monoprotic acids and monobasic bases (one H+ ion and one OH- ion), the following equation can be used: molarityacid x volumeacid = molaritybase x volumebase Using the information from the previous problem, and plugging in to the equation above yields: ? M x 20 mL = .25 M x 25 mL Solving for ? Yields: ?M = (.25 M x 25mL)/20mL = M

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**Titration Shortcut II -Sample**

50 mL of NaOH is titrated with 25 mL of .5M H2SO4 . What is the Molarity of the NaOH solution? Write the balanced equation: 2 NaOH + 1 H2SO4 Na2SO H2O Plug into the equation: Coeff.acid x Macid x Vacid = Mbase xVbase x Coeff.base .5M x 50 ml / 1 = ? M x 25 mL / 2 Solve for ?Mbase = 1.0 M

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**Macid x Vacid /Coeffacid = Mbase xVbase / Coeff.base**

Titration Shortcut II For titrations between any acids and bases the following equation can be used: Macid x Vacid /Coeffacid = Mbase xVbase / Coeff.base Use the above to solve the following: 40 mL of NaOH is titrated with 30 mL of .5M H2SO4 . What is the Molarity of the NaOH solution?

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