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Question: What is the pH of the following solutions? A.\({10^{ - 3}}M\,HCL\) B.\(0.0001M\,NaOH\) C.\(0.0...

What is the pH of the following solutions?
A.10−3M HCL{10^{ - 3}}M\,HCL
B.0.0001M NaOH0.0001M\,NaOH
C.0.0001M H2SO40.0001M\,{H_2}S{O_4}

Explanation

Solution

pH ('potential of hydrogen' or 'power of hydrogen') is a chemical scale for determining the acidity or basicity of an aqueous solution. Acidic solutions (those containing a larger concentration of H+{H^ + }ions) have a lower pH than basic or alkaline solutions.

Complete answer:
The pH of a chemical solution indicates how acidic or basic it is. A pH value of seven is considered neutral, less than seven acidic, and greater than seven basic on the pH scale, which ranges from 00 to1414
Acids and bases are defined in a variety of ways, but pH relates exclusively to hydrogen ion concentration and is used to describe aqueous (water-based) solutions. When water dissociates, a hydrogen ion and a hydroxide are produced.
H2O↔ H++OH−{H_2}O \leftrightarrow \,{H^ + } + O{H^ - }
Remember that [ ] stands for molarity, M, when calculating pH. The molarity of a solution is measured in moles of solute per litre of solution. If the concentration is given in a unit other than moles (mass percent, molality, etc. ), convert it to molarity before applying the pH formula.
pH can be calculated as:
pH = −log  10[H  +]pH{\text{ }} = {\text{ }} - log{\;_{10}}[H{\;^ + }]
[H  +] = 10  −pH[H{\;^ + }]{\text{ }} = {\text{ }}10{\;^{ - pH}}
Now, lets come to problem
(a)
HClHCl is a strong electrolyte that is fully ionised.
HCl⇌H++Cl−HCl \rightleftharpoons {H^ + } + C{l^ - }
So, [H+]=10−3 M[{H^ + }] = {10^{ - 3}}\,M
pH=−log⁡[H+] =−log⁡ 10−3=3 = - \log [{H^ + }]\, = - \log \,{10^{ - 3}} = 3
(b) NaOHNaOH is a strong electrolyte that is fully ionized.
NaOH⇌Na++OH−NaOH \rightleftharpoons N{a^ + } + O{H^ - }
So, [OH+]=0.0001M=10−4M[O{H^ + }] = 0.0001M = {10^{ - 4}}M
pOH=−log⁡ 10−4=4pOH = - \log \,{10^{ - 4}} = 4
(c). H2SO4{H_2}S{O_4} is a strong electrolyte that is fully ionised
H2SO4⇌2H++SO42−{H_2}S{O_4} \rightleftharpoons 2{H^ + } + SO_4^{2 - }
So, [H+]=2×10−4 M[{H^ + }] = 2 \times {10^{ - 4}}\,M
pH=−log⁡[H+]=3.70pH = - \log [{H^ + }] = 3.70

Note:
A glass electrode with a pH metre, or a color-changing indicator, can be used to determine the pH of aqueous solutions. pH measurements are useful in chemistry, agronomy, medicine, water treatment, and a variety of other fields.