Solveeit Logo
Login

Question

Question: The pH of the solution \[0.1M\]\[{H_2}S{O_4}50(ml)\] + \[0.4M\]\[HCl50(ml)\] is: A. 5 B. 7 C. ...

The pH of the solution 0.1M$$$${H_2}S{O_4}50(ml) + 0.4M$$$$HCl50(ml) is:
A. 5
B. 7
C. 12
D. None of the above

Explanation

Solution

Molarity of the solution is defined as the number of moles of solute per litre of the solution. It can also be expressed in terms of millimoles as the number of millimoles of solute per milliliter of solution.

Complete step by step answer:
Molarity of the solution can be expressed as,
M=nVM = \dfrac{n}{V}
Where,
n-moles of solute(mol/m.mol)
V-volume of solution (l/ml)
0.1M0.1Mof H2SO4{H_2}S{O_4} is present in 50(ml)50(ml) the solution. Thus, millimoles present in H2SO4{H_2}S{O_4}can be calculated by using the above Molarity formula.
Millimoles of H2SO4{H_2}S{O_4}=0.1×500.1 \times 50
=5= 5 millimoles
Thus, 5 millimoles are present in 0.1M0.1Mof H2SO4{H_2}S{O_4} solution.
Similarly, 0.4M0.4Mof HClHCl is present in 50(ml)50(ml) the solution. Thus, millimoles present in HClHCl can be calculated by using the above Molarity formula.
Millimoles of HClHCl=0.4×500.4 \times 50
=20= 20 millimoles
Thus, 20 millimoles are present in 0.4M0.4M of HClHCl solution.
Total volume present in the solution = Volume of H2SO4{H_2}S{O_4}+Volume of HClHCl
=50+50= 50 + 50
=100mL= 100mL
Total millimoles present in 0.1MH2SO450(mL)  +  0.4M  HCl50(mL)=5+200.1M{{ }}{{{H}}_2}S{O_4}{{ 50(mL)\; + \; 0}}{{.4M\; HCl 50(mL) = 5 + 20}}
=25millimoles= 25{{ millimoles}}
Molarity of 0.1MH2SO450(mL)  +  0.4MHCl50(mL)0.1M{{ }}{{{H}}_2}S{O_4}{{ 50(mL) \; +\; 0}}{{.4M HCl 50(mL)}} can be calculated as,
M=25m.mol100mlM = \dfrac{{25m.mol}}{{100ml}}
M=0.25MM = 0.25M
Thus, Molarity of 0.1MH2SO450(mL)  +  0.4MHCl50(mL)=0.25M0.1M{{ }}{{{H}}_2}S{O_4}{{ 50(mL) \; + \;0}}{{.4M HCl 50(mL) = 0}}{{.25M}} which is nothing but the concentration of free hydrogen ion in a solution. This is because HClHCl and H2SO4{H_2}S{O_4} is a strong acid and thus, it contains free hydrogen ions.
It is known that pH is the negative logarithm of the concentration of hydrogen ion which it contains. This can be written as follows,
pH=−log⁡(H+)pH = - \log ({H^ + })
Thus, the pH for 0.1M0.1M H2SO450(ml){H_2}S{O_4}50(ml) + 0.4M0.4M HCl50(ml)HCl50(ml) can be calculated as,
pH=−log⁡(0.25)pH = - \log (0.25)
pH=0.6020pH = 0.6020
Thus, the pH for 0.1M0.1M H2SO450(ml){H_2}S{O_4}50(ml) + 0.4M0.4M HCl50(ml)HCl50(ml) is 0.60200.6020. And it is not present in any options.

Thus, the correct answer is option D.

Note: Similar to pH, the concentration of hydroxyl ion contains cal be calculated as, pOH=−log⁡[OH−]pOH = - \log [O{H^ - }]
pOH is the negative logarithm of hydroxyl ion present in the solution. The relation between pH and pOH can be written as; pOH+pH=14pOH + pH = 14.Thus, if the pOH of a solution is known, then the pH of that solution can be calculated.