Question

Equal volumes of two solutions pH=2 and pH=4 are mixed together.Calculate the pH of the resulting solution?

Answer 1

note that :-

✔️✔️[H+] for solution with pH = 2 is 10-2 M
✔️✔️[H+] for solution with pH = 4 is 10-4 M

equal volumes of the given solutions are mixed so we know that the concentration will be halved.

[H+] =10-2 +10-42   
      
=10-2 (1+10-2)2 
        
=1.01 x 10-22   
      
=0.505 x 10-2 M

pH = -log[H+]   
    
= -log (0.505 x 10-2) 
      
=2.29


➖➖➖➖Better method ➖➖➖➖


✔️✔️let x be the vol. of both sol.

M1V1 +M2V2=Mnet Vnet

10-2 x *+10-4 X x= Mnet 2x

x(10-2 + 10-4) =Mnet 2x

10-2+10-4/2 = Mnet

Mnet= 5.05 X 10-3

take -log

=2.2967

pH=2.2967

or pH=2.3


$\huge\bf{\underline {\mathfrak{thank \: you :)}}}$

Answer 2

Answer:

The pH of the resulting solution is equal to 2.3053.

Explanation:

Given: pH of one solution $=2$

pH of second solution $=4$

We know that, $pH=-log[H^{+}]$

Therefore, $[H^{+}]=10^{-pH}$

So, the concentration of [H⁺] in first solution $=10^{-2}$

Concentration of [H⁺] in 2nd solution $=10^{-4}$

When these two solutions mixed together, concentration of [H⁺] will be:

$[H^{+}]=\frac{10^{-2}+10^{-4}}{2}$

Therefore, pH of resulting solution will be:

$pH=-log[H^{+}]$

$pH=-log(\frac{10^{-2}+10^{-4}}{2} )$

$pH=-[log(10^{-2}+10^{-4})-log2]$

$pH=-log(10^{-2})(1+10^{-2})+log2\\pH=-log(10^{-2})+log(1+0.01)+log2\\pH=-log(10^{-2})+log(1.01)+log2$

$pH=2log10+log1.01+log2$

$pH=2+0.0043+0.3010\\pH=2.3053$

Therefore, the pH of the resulting solution after mixing will be equal to 2.3053.