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Question: What is the \(pH\)of \(0.05M\) of \(HCl\)?...

What is the pHpHof 0.05M0.05M of HClHCl?

Explanation

Solution

We know that pHpH of a solution is defined as the concentration of hydrogen ions in the solution. pH scale ranges from value between 00 to 77. pH is also defined as the negative of the logarithm base 1010 of the molar concentration of hydrogen ions in the solution

Complete answer:
We know that pH of a solution is defined as the concentration of hydrogen ions in the solution. pHpH scale ranges from value between 00 to 77. Lower the value of pH, higher is the acidity of the solution. Higher the value of pH more basic will be the solution. pH is also defined as the negative of the logarithm base 1010 of the molar concentration of hydrogen ions in the solution. For bases, we can calculate pOHpOH, which is defined as the negative of the logarithm base 1010 of the molar concentration of OHO{H^ - } ions in the solution. For pH, we can therefore write
pH=log10[H+]pH = - {\log _{10}}[{H^ + }] ……(1)\left( 1 \right)
Here we are given hydrochloric acid. The molecular formula of hydrochloric acid is HClHCl . The molar concentration of hydrochloric acid given is 0.05M0.05M.
When HClHCl is dissolved in water, it is given one H+{H^ + } ion. So, 0.05M0.05M will give 1×0.05M1 \times 0.05M hydrogen ions which can be written as
[H+]=0.05M[{H^ + }] = 0.05M
[H+]=[{H^ + }] = Concentration of hydrogen ions
Putting this value in equation (1)\left( 1 \right) , we get
pH=log10[H+]pH = - {\log _{10}}[{H^ + }]
pH=log[0.05]\Rightarrow pH = - \log [0.05]…… (2)\left( 2 \right)
0.050.05 can be written as 5×1025 \times {10^{ - 2}}. Substituting this value in (2)\left( 2 \right)
pH=log[5×102]\Rightarrow pH = - \log [5 \times {10^{ - 2}}]
pH=1.3\Rightarrow pH = 1.3

Thus, pHpH is 1.31.3.

Note: It is important to note pHpH of a neutral is 77. All the values of pHpH which are less than 77 are acidic in nature. Lesser is the value of pHpH, more acidic in nature. Values of pHpH greater than 77 are basic in nature. Also pH+pOH=14pH + pOH = 14