Question

The ph of a solution obtained by mixing 10ml of 0.45M HCL and 40 ml of 0.1 M NaOH is

Answer 1

We need to find concentration of resulting solution So we need to use the formula
$m = \frac{v1m1 - v2m2}{v1 + v2}$
where m is molarity of resulting solution, V1 and V2 are volumes of acid and base, m1 and m2 are molarities of them.
m= 10×0.45-40×0.1/(40+10)= 0.5/50= 0.01
As pH is negative algorithm to H+ concentration ,
we know that acid has neutralised the base so
$- log(0.01) = - log( {10}^{ - 2} ) = 2$
Hence pH of the resulting solution is 2.

Answer 2

Hey dear,

● Answer -
pH = 2

● Explanation -
# Given -
C1 = 0.45 M
C2 = 0.1 M
V1 = 10 ml
V2 = 40 ml

# Solution -
Final concentration of the solution is calculated by -
C = (C1V1 - C2V2) / (V1 + V2)
C = (0.45×10 - 0.1×40) / (10 + 40)
C = 0.01 M

pH of resultant solution is given by -
pH = -log(conc.)
pH = -log(0.01)
pH = 2

Therefore, pH of resultant solution is 2.

Hope this helps you..