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Question: 100ml of \[0.1M\] \[HCl\] is mixed with \(100ml\) of \[0.01M\]\[HCl\] .The \[pH\] of the resulting s...

100ml of 0.1M0.1M HClHCl is mixed with 100ml100ml of 0.01M$$$$HCl .The pHpH of the resulting solution is:
A. 2.02.0
B. 1.01.0
C. 1.261.26
D. None of these

Explanation

Solution

The term “pHpH” is an abbreviation for the “potential of hydrogen.” pHpH is a unit of measurement which represents the concentration of hydrogen ions in a solution.

Complete step by step answer:
To calculate the pHpH of an aqueous solution you need to know the concentration of the hydronium ion in moles per liter (molarity). The pHpH is then calculated using the expression:
pH=log[H3O]+pH = - \log {[{H_3}O]^ + }
Calculating the Hydronium Ion Concentration from pHpH
The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pHpH .
[H3O]+=10pH{[{H_3}O]^ + } = {10^{ - pH}}
Or
[H3O]+=antilog(pH){[{H_3}O]^ + } = anti{\log ^{( - pH)}}
Calculating pOHpOH
To calculate the pOHpOH of a solution you need to know the concentration of the hydroxide ion in moles per liter (molarity). The pOHpOH is then calculated using the expression:
pOH=log[OH]pOH = - \log {[OH]^ - }
Calculating the Hydroxide Ion Concentration from pOHpOH
The hydroxide ion concentration can be found from the pOHpOH by the reverse mathematical operation employed to find the pOHpOH .
[OH]=10pOH{[OH]^ - } = {10^{ - pOH}}
Or
[OH]=antilog(pOH){[OH]^ - } = anti\log ( - pOH)
Relationship Between and pOHpOH .
The pHpH and pOHpOH of a water solution at 25^\circ C are related by the following equation.
pH+pOH=14pH + pOH = 14
Step by step solution: When you add two different concentrations of the same strong acid, the pH contribution will be from both concentrations.
M1=0.1M V1=100ml M2=0.01M V2=100ml M_1 = 0.1M \\\ V_1 = 100ml \\\ M_2 = 0.01M \\\ V_2 = 100ml
Final volume of solution is 100+100=200ml100 + 100 = 200ml
The final concentration of mixture will be calculated as
MV=M1V1+M2V2MV = M_1V_1 + M_2V_2
= M(100+100)=0.1×100+0.01×100MM(100 + 100) = 0.1 \times 100 + 0.01 \times 100M
M=200(10+1)=0.055MM = 200(10 + 1) = 0.055M
pH=log[H+]pH = - \log [{H^ + }]
=log(0.055) - \log (0.055)
=1.2591.261.259 \sim 1.26

So, the correct answer is Option C.

Note: The pHpH scale describes the acidity of the solution: acidic, neutral, or basic A solution with a pHpH less than 7 is an acid, exactly 7 is a neutral solution, and above 7 is a base. Bases have less hydrogen ions but more hydroxide ions, represented by the pOHpOH or “potential of hydroxide ions.”

AcidicNeutralBasic
Less than 77Greater than 7