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Wize High School Grade 12 Chemistry Textbook > Acids and Bases

Applications of Acid and Base Equations

Strength of Acids and Bases Previous Section
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Example: pH of Acids and Bases

Part 1: What is the pH of a 0.5 M HCl solution?

HCl is a (strong/weak acid)
strong
It dissociates (completely/incompletely)
completely

Strong acids COMPLETELY dissociate so we can write the reaction: HCl → H+ + Cl-
0.5M
Since the equation is balanced, we can see that there is a 1:1 molar ratio between HCl and H+ so the concentration of H+ will actually be the same!
[H+] =0.5M
plug that into the pH equation: pH=-log[H+]
pH=-log[0.5]
pH=0.30


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Part 2: What is the pH of a 1.0 M CH3COOH solution? Ka = 1.8 x 10-5

This is an (acid/base)
acid
It is (weak/strong)
weak
Does this type of acid dissociate into ions completely/incompletely?
incompletely

CH3COOH is not one of the strong acids we had to memorize, so we automatically know that it is a weak acid! Weak acids have INCOMPLETE dissociation, so we will see an equilibrium in the equation:

CH3COOH(aq) + H2O(l) ⇌ CH3COO- + H3O+(aq)
1.0M ? need this to find pH!

Write out an ICE Table:
I 1M / 0 0
C -x +x +x
E 1-x x x

Write out the Ka expression:

Ka=[CH3COO−][H3O+][CH3COOH]Ka=\frac{\left[CH_3COO^-\right]\left[H_3O^+\right]}{\left[CH_3COOH\right]}

Plug in the Ka value we were given and plug in the equilibrium concentrations from our ICE table:

1.8x10−5=[x][x][1−x]1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[1-x\right]}

Let's check to see if the denomenator has to stay [1-x] (and we would have to use the quadratic equation) OR we might be able to simplify it to ~1.
When the denominator is in the form of y-x, we can check to see if y/K > 400 and if it is, then we can make the assumption and simplify.

Let's see.... 1/1.8x10-5= 55 555 which is much larger than 400! This means that our denominator was [1-x] but we can simplify it to [1] since y is so large and K is so small so the x won't make a difference.

1.8x10−5=x211.8x10^{-5}=\frac{x^2}{1}

Take the square root of both sides:
4.24x10-3=x

If we look back at our ICE table, x=[H+]

so now we can solve for pH:
pH=-log[H+]
pH=-log[4.24x10-3]
pH=2.37 !


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Part 3: What is the pH of a solution prepared by dissolving 0.10 mol of Ba(OH)2 in 1.0 L of pure water?

Is this an acid or base?
base
Is this a strong one or weak one?
Strong
Complete/incomplete dissociation?
Complete

Since we know it's a strong base, that means we know it will COMPLETELY dissociate so we can write out the chemical reaction:
Ba(OH)2 → Ba2+ + 2OH- (Always double check your equation is balanced!!)
Here we are given n=0.10 mol, V=1L. We can use the equation n=cv to solve for c:
c=n/v
c=0.10mol/1L
c=0.1M
0.1M 0.2M
(Since Ba(OH)2 has a concentration of 0.1M, OH- must have a concentration of 0.2M since there is a 1:2 molar ratio!)

[OH-]=0.2M
pOH=-log[OH-]
pOH=-log[0.2]
pOH=0.70
We are asked to solve for pH so use the equation: pH + pOH = 14
pH=14-pOH
pH=14-0.70
pH=13.3 (This logically makes sense since we would expect a base to have a very high pH)

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Part 4: Many liquid household cleaners contain ammonia. The concentration of ammonia (NH3) in these products is usually around 5.0 M. What is the pH of such a cleaning solution? (Kb NH3 = 1.8 x 10-5)

Is this an acid or a base?
Base
Is it strong or weak?
Weak
Complete or incomplete dissociation?
Incomplete

NH3 is a weak base so that means it INCOMPLETELY dissociates:
NH3 + H2O ⇌ NH4+ + OH-
5M
Here we will need to make an ICE table, so we can again plug in the equilibrium concentrations into our Ka expression, solve for [OH-] then pOH, then pH !
I 5M / 0 0
C -x +x +x
E 5-x x x

Now write the Kb expression out:

Kb=[NH4+][OH−][NH3]Kb=\frac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}

Plug in the Kb value we were given and the equilibrium values from our ICE Table:
1.8x10−5=[x][x][5−x]1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[5-x\right]}

Check to see if we can make the simplification of the denominator from [5-x] to ~5.
y/K >400
5/1.8x10-5=277 778. which is much greater than 400 so we can make the simplifcation!

1.8x10−5=[x][x][5]1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[5\right]}
1.8x10-5(5)=x2
9x10-5=x2
Take the square root of both sides:
9.49x10-3=x
x=[OH-] (see the ICE Table)
Solve for pOH:
pOH=-log[OH-]
pOH=-log[9.49x10-3]
pOH=2.02

The question is asking us to solve for pH so use pH + pOH=14
pH=14 - pOH
pH=14-2.02
pH=11.98 !

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The digestion of some high-protein foods such as red meat, beans, lentils, and shell-fish, releases uric acid, HC5H3N4O3(aq), which ionizes into hydrogen ions, H+(aq) and urate ions, C5H3N4O3-(aq) in the bloodstream. People whose kidneys do not function properly cannot excrete urate ion sufficiently quickly, leading to an increased concentration of urate in the blood. This sometimes leads to a painful form of arthritis known as gout, characterized by the formation of tiny needle-like crystals of sodium urate, NaC5H3N4O3(aq), in joints and tissues according to the equation:
NaC5H3N4O3(s) ⇌ Na+(aq) + C5H3N4O3-(aq)

a) Suppose you are a nutritionist. What advice could you give to your patients who suffer from gout? Explain why the advice would be effective [2]


b) Many women take calcium supplements on a daily basis to prevent the loss of bone mass (a condition known as osteoporosis). If a woman suffering from osteoporosis has gout too, she may also develop kidney stones (which can consist of calcium urate). Write a chemical equilibrium equation for this reaction and explain why it happens. [2]