To determine the pH of the resultant solution, we start by calculating the concentrations of
OH−ions in the NaOH solution and
H+ions in the HCl solution.
Given that the pH of the NaOH solution is 10 , we can find the concentration of
OH−ions as follows:
pH of NaOH solution
=10Since
pOH=14−pH, we get:
pOH=14−10=4 The concentration of
OH−ions in the NaOH solution can be calculated using:
[OH−]=10−pOH=10−4MSo, the moles of
OH−ions in 300 ml of the solution will be:
Volume (in liters)
=0.3LMoles of
OH−ions
=0.3L×10−4M=3×10−5‌molesNext, we consider the HCl solution with a pH of 4 :
pH=4The concentration of
H+ions in the HCl solution is:
[H+]=10−pH=10−4MMoles of
H+ions in 200 ml of solution will be:
‌ Volume ‌(‌ in liters ‌)=0.2L Moles of
H+ions
=0.2L×10−4M=2×10−5‌molesWhen the solutions are mixed, hydroxide ions
(OH−)will react with hydrogen ions
(H+)to form water:
OH−+H+⟶H2OThe moles of
OH−and
H+present are:
Moles of
OH−=3×10−5Moles of
H+=2×10−5 Since
H+will neutralize an equal number of
OH−ions, the remaining moles of
OH−ions will be:
3×10−5−2×10−5=1×10−5‌moles of
OH−ions
The total volume of the resultant solution is:
300‌ml+200‌ml=500‌ml(0.5L) The concentration of remaining
OH−ions in the resultant solution is:
[OH−]=‌| 1×10−5‌ moles ‌ |
| 0.5L |
=2×10−5MNow we calculate the pOH of the resultant solution:
pOH=−log10(2×10−5) We can use the logarithmic properties for simplification:
‌log10(2×10−5)=log10(2)+log10(10−5)‌log10(2)≈0.301Therefore,
log10(2×10−5)=0.301−5=−4.699Hence,
pOH=4.699Given
pH+pOH=14, we find the pH as follows:
pH=14−pOH=14−4.699=9.301Therefore, the pH of the resultant solution is 9.301 .
The correct answer is:
Option C: 9.301