Ph of Acids and Bases - Definition, Examples, Limitations, FAQs

The theory of pH was introduced by Danish chemist Soren Peder Lauritz Sorensen in the year of 1909. Sorensen developed the pH scale to measure the acidity or Basicity of a solution more accurately. The scale that he discovered to measure the acidity and basicity of the compound expresses the concentration of hydrogen ions in the solution, the lower pH is the indicator of higher Hydrogen ion concentration and then that of higher pH indicates the lower hydrogen ion concentration, and a very basic namely Alkaline Solution.

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This Story also Contains

1. Ph Of Acids And Bases
2. Some Solved Examples
3. Summary

Ph Of Acids And Bases

what is pH

pH is also referred to as the potential or power of hydrogen. Mathematically, it can be represented as follows:

$\mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$

If the solution is neutral, then:
K_w = [H₃O⁺][OH^-]
From the ionic product of water, we know:
K_w = 10^-14
[H₃O⁺] = [OH^-] = x (since solution is neutral)
Thus, 10^-14 = K_w = x²
x = 10^-7
Now, [H₃O⁺] = 10^-7
Thus, pH = - log₁₀(H₃O⁺) = - log₁₀(10^-7) = 7

For Acidic solutions: For Basic solutions:

For acidic solutions, we must have [H₃O⁺] > [OH^-] For basic solutions, we must have [H₃O⁺] < [OH^-]
Thus, [H₃O⁺] > 10^-7 Thus, [H₃O⁺] < 10^-7
Thus, [H₃O⁺] for acids can be 10^-6, 10^-5, 10^-4, etc. Thus, [H₃O⁺] for basics can be 10^-8, 10^-9, 10^-10, etc.
Thus, pH of acids can be 6, 5, 4, etc. Thus, pH of basics can be 8, 9, 10, 11, etc.
Hence, pH of acidic solutions is less than 7 Hence, pH of basic solutions is greater than 7

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pH depends upon the temperature

We know from the ionic product of water that at 63⁰C, the value of K_w = 10^-13.
For a neutral solution, we know the:
$\begin{aligned} & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]} \\ & \Rightarrow \mathrm{K}_{\mathrm{w}}=\mathrm{x}^2 \\ & \Rightarrow \mathrm{x}=\sqrt{10^{-13}}=10^{-6.5} \\ & \Rightarrow\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-6.5} \\ & \Rightarrow \mathrm{pH}=-\log _{10}\left(10^{-6.5}\right)=6.5\end{aligned}$

Hence, pH depends upon the temperature

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pH of Strong Acids

Strong acids are those acids that dissociate completely in solutions. For example:

• 2 x 10^-3 M HNO₃
  Since HNO₃ is a strong acid, thus it will dissociate completely into H⁺ and OH^- ions as follows:
  $\begin{aligned} & \mathrm{HNO}_3 \rightarrow \mathrm{H}^{+}+\mathrm{NO}_3^{-} \\ & \text {Thus, }\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M} \quad \text { (given) } \\ & \Rightarrow \mathrm{pH}=-\log _{10}\left(2 \times 10^{-3}\right) \\ & \Rightarrow \mathrm{pH}=-\log _{10}(2)-\log _{10}\left(10^{-3}\right) \\ & \Rightarrow \mathrm{pH}=-0.30+3=2.7\end{aligned}$
  Thus, the pH of HNO₃ is 2.7
• 10^-4 M H₂SO₄

Since H₂SO₄ is a strong acid, thus it will dissociate completely into H⁺ and OH^- ions as follows:

$\begin{aligned} & \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-} \\ & \text { Thus, }\left[\mathrm{H}^{+}\right]=2 \times 10^{-4} \mathrm{M} \\ & \Rightarrow \mathrm{pH}=-\log _{10}\left(2 \times 10^{-4}\right) \\ & \Rightarrow \mathrm{pH}=-\log _{10}(2)-\log _{10}\left(10^{-4}\right) \\ & \Rightarrow \mathrm{pH}=-0.30+4=3.7\end{aligned}$
Thus, the pH of H₂SO₄ is 3.7

NOTE: If molarity(N) of solution is not given but normality(N) is given, then molarity can be calculated using the following formula:

N = M x n
where n is the number of moles

Related topics link,

• Solubility Product Constant
• Propertries of Acids and Bases
• Neutralization Reaction

the pH of Weak Acids

Weak acids are those acids that dissociate partially in solutions. For example:

• 8 M HA (K_a =2 x 10^-8)
  The chemical equation for the dissociation of weak acid HA is as follows:
  $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$
  Initial: 8M 0 0
  Equil: 8 - 8? 8? 8?
  The equilibrium constant K_a for the weak acid is given as follows:
  $\begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{8 \alpha .8 \alpha}{8(1-\alpha)}=\frac{8 \alpha^2}{1-\alpha} \\ & \mathrm{K}_{\mathrm{a}}=8 \alpha^2 \quad(\text { as }(1-\alpha) \approx 1) \\ & \text { Thus, } \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{8}}=\sqrt{\frac{2 \times 10^{-8}}{8}}=\sqrt{\frac{10^{-8}}{4}}=0.5 \times 10^{-4} \\ & {\left[\mathrm{H}^{+}\right]=8 \times 0.5 \times 10^{-4}=4 \times 10^{-4}} \\ & \mathrm{pH}=-\log _{10} 4+4 \\ & \mathrm{pH}=-0.60+4=3.4\end{aligned}$
  Thus, the pH of this given acid = 3.4
• 0.002N CH₃COOH(? = 0.02)
  The chemical equation for the dissociation of CH₃COOH is as follows:
  $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
  Initial: c 0 0
  Equil: c - c? c? c?
  The equilibrium constant K_a for the weak acid is given as follows:
  $
  \begin{aligned}
  & \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COOH}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{\mathrm{c} \alpha \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=\frac{\mathrm{c} \alpha^2}{1-\alpha} \\
  & \mathrm{K}_{\mathrm{a}}=\mathrm{c} \alpha^2 \quad(\text { as }(1-\alpha) \approx 1)
  \end{aligned}
  $
  Now, as we have given :
  $
  \begin{aligned}
  & \mathrm{c}=0.002 \mathrm{~N} \text { or } 0.002 \mathrm{M} \quad(\text { Normality }=\text { Molarity, as } \mathrm{n} \text { factor }=1) \\
  & \alpha=\frac{2}{100}=0.02
  \end{aligned}
  $
  Thus, $\left[\mathrm{H}^{+}\right]=0.002 \times 0.02=4 \times 10^{-5}$
  $
  \begin{aligned}
  & \mathrm{pH}=-\log _{10}\left(4 \times 10^{-5}\right) \\
  & \mathrm{pH}=5-\log 4=4.4
  \end{aligned}
  $
  Thus, the pH of acetic acid = 4.4

Recommended topic video on (pH Of Acids And Bases)

Some Solved Examples

Example.1

1. Which one of the following statements is not true?

1)The conjugate base of $\mathrm{H}_2 \mathrm{PO}_4{ }^{-}$is $\mathrm{HPO}_4{ }^{2-}$
2) $p H+p O H=14$ for all aqueous solutions.
3) (correct) The pH of $1 \times 10^{-8} \mathrm{MHCL}$ is 8
4)96,500 coulombs of electricity when passed through a $\mathrm{CuSO}_4$ solution deposits 1 gram equivalent of copper at the cathode.

Solution

Value of pH -An acidic solution has $p H<7$, The basic solution has $p H>7$, Neutral solution has $p H=7. p H_{\text {acid cannot exceed } 7 \text { Here we should also consider }}\left[\mathrm{H}^{+}\right]$that comes from $H_2 \mathrm{O}$ Now $\left[\mathrm{H}^{+}\right]=\left[\mathrm{H}^{+}\right]_{\text {from } \mathrm{HCl}}+\left[\mathrm{H}^{+}\right]_{\text {from } \mathrm{H}_2 \mathrm{O}}$
$
\begin{aligned}
{[\mathrm{H}]^{+} } & =10^{-8}+10^{-7} \\
{[\mathrm{H}]^{+} } & =10^{-8}+10 \times 10^{-8} \\
{[\mathrm{H}]^{+} } & =11 \times 10^{-8}
\end{aligned}
$
$
\therefore p H=-\log \left(11 \times 10^{-8}\right)=6.9587
$

Hence, the answer is the option (3).

Example.2

2. An alkali is titrated against an acid with methyl orange as an indicator, which of the following is a correct combination?

1)Base Acid End Point

Strong Strong Pink to colorless

2)Base Acid End Point

Weak Strong Colourless to pink

3)Base Acid End Point

Strong Strong Pinkish red to yellow

4) (correct)Base Acid End Point

Weak Strong Yellow to pinkish red

Solution

As we learned in

Value of p(H) -

Acidic solution has p(H) < 7

Basic solution has p(H) > 7

Neutral solution has p(H) = 7

Methyl orange changes its color in acidic medium when pH is around 3 to 5 and its colour is yellow in basic medium and red in acidic medium so it is used to titrate weak base with strong acid.

Hence, the answer is an option (4).

Example.3

3. Among the following acids which have the lowest $p K_a$ value?

1) $\mathrm{CH}_3 \mathrm{COOH}$
2) $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{COOH}$
3) (correct) HCOOH
4) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$

Solution

The higher the $p K_a$ value, the weaker is the acid. Hence, stronger acid has lower $p K_a$ value.

In the given options, HCOOH is the strongest acid.

Hence, the answer is the option (3).

Example.4

4.Hydrogen ion concentration in mol/L in a solution of $p H=5.4$ will be

1)$3.98 \times 10^8$

2)$3.88 \times 10^6$

3)$3.68 \times 10^{-6}$

4) (correct)$3.98 \times 10^{-6}$

Solution

The pH scale -

Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the pH scale.

The pH of a solution is defined as the negative logarithm to base 10 of the activity of hydrogen ion

$p H=-\log \left[H^{+}\right]$

$\left[H^{+}\right]=\operatorname{antilog}(-p H)=\operatorname{antilog}(-5.4)=3.98 \times 10^{-6}$

Hence, the answer is the option (4).

Example.5

5.How many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?

1) (correct)9.0 L

2)0.1 L

3)0.9 L

4)2.0 L

Solution

If molarity(N) of solution is not given but normality(N) is given, then molarity can be calculated using the following formula:

N = M x n
where n is the number of moles

Initial PH$=2=1 \log \left[H^{+}\right]_2$

$\left[H^{+}\right]_1 \times 1=\left[H^{+}\right]_2 \times V$

$10^{-1} \times 1=10^{-2} \times V$

V=10L

Added water =10-1=9L

Hence, the answer is the option (1).

Summary

PH is describe as the negative logarithm (base 10) of the hydrogen ion concentration. Acids are substances that donate hydrogen ions (protons) to any solution. They have a pH of less than 7. Strong Acids are those that are completely dissociated in water, such as hydrochloric acid (HCl) and sulfuric acid (H₂SO₄). They have a low pH, which is close to 0. Weak acids such as acetic acid (CH3COOH) partially dissociate in water. They have a pH closer to 7 but still less than 7 . Examples: Lemon juice, vinegar, and battery acid. Bases are substances that accept hydrogen ions or donate hydroxide ions (OH-) in solution. They have a pH greater than 7. Strong bases such as sodium hydroxide (NaOH) and potassium hydroxide (KOH) are completely dissociated in water. They have a high pH, often close to 14. Weak Bases are partially dissociated in water, such as ammonia (NH3). They have a pH almost equal to 7 but still more than 7.

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