To determine the $$\mathrm{pH}$$ of the resultant solution, we start by calculating the concentrations of $$\mathrm{OH^-}$$ ions in the $$\mathrm{NaOH}$$ solution and $$\mathrm{H^+}$$ ions in the $$\mathrm{HCl}$$ solution.

Given that the $$\mathrm{pH}$$ of the $$\mathrm{NaOH}$$ solution is 10, we can find the concentration of $$\mathrm{OH^-}$$ ions as follows:

$$\mathrm{pH}$$ of $$\mathrm{NaOH}$$ solution = 10

Since $$\mathrm{pOH} = 14 - \mathrm{pH}$$, we get:

$$\mathrm{pOH} = 14 - 10 = 4$$

The concentration of $$\mathrm{OH^-}$$ ions in the $$\mathrm{NaOH}$$ solution can be calculated using:

$$[\mathrm{OH^-}] = 10^{-\mathrm{pOH}} = 10^{-4} \, \mathrm{M}$$

So, the moles of $$\mathrm{OH^-}$$ ions in $$300 \, \mathrm{ml}$$ of the solution will be:

Volume (in liters) = $$0.3$$ L

Moles of $$\mathrm{OH^-}$$ ions = $$0.3 \, \mathrm{L} \times 10^{-4} \, \mathrm{M} = 3 \times 10^{-5} \, \mathrm{moles}$$

Next, we consider the $$\mathrm{HCl}$$ solution with a $$\mathrm{pH}$$ of 4:

$$\mathrm{pH} = 4$$

The concentration of $$\mathrm{H^+}$$ ions in the $$\mathrm{HCl}$$ solution is:

$$[\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-4} \, \mathrm{M}$$

Moles of $$\mathrm{H^+}$$ ions in $$200 \, \mathrm{ml}$$ of solution will be:

Volume (in liters) = $$0.2$$ L

Moles of $$\mathrm{H^+}$$ ions = $$0.2 \, \mathrm{L} \times 10^{-4} \, \mathrm{M} = 2 \times 10^{-5} \, \mathrm{moles}$$

When the solutions are mixed, hydroxide ions ($$OH^-$$) will react with hydrogen ions ($$\mathrm{H^+}$$) to form water:

$$\mathrm{OH^- + H^+ \rightarrow H_2O}$$

The moles of $$\mathrm{OH^-}$$ and $$\mathrm{H^+}$$ present are:

Moles of $$\mathrm{OH^-}$$ = $$3 \times 10^{-5}$$

Moles of $$\mathrm{H^+}$$ = $$2 \times 10^{-5}$$

Since $$\mathrm{H^+}$$ will neutralize an equal number of $$\mathrm{OH^-}$$ ions, the remaining moles of $$\mathrm{OH^-}$$ ions will be:

$$3 \times 10^{-5} - 2 \times 10^{-5} = 1 \times 10^{-5} \, \mathrm{moles}$$ of $$\mathrm{OH^-}$$ ions

The total volume of the resultant solution is:

$$300 \, \mathrm{ml} + 200 \, \mathrm{ml} = 500 \, \mathrm{ml} \, (0.5 \, \mathrm{L})$$

The concentration of remaining $$\mathrm{OH^-}$$ ions in the resultant solution is:

$$[\mathrm{OH^-}] = \frac{1 \times 10^{-5} \, \mathrm{moles}}{0.5 \, \mathrm{L}} = 2 \times 10^{-5} \, \mathrm{M}$$

Now we calculate the $$\mathrm{pOH}$$ of the resultant solution:

$$\mathrm{pOH} = -\log_{10}(2 \times 10^{-5})$$

We can use the logarithmic properties for simplification:

$$\log_{10}(2 \times 10^{-5}) = \log_{10}(2) + \log_{10}(10^{-5})$$

$$\log_{10}(2) \approx 0.301$$

Therefore,

$$\log_{10}(2 \times 10^{-5}) = 0.301 - 5 = -4.699$$

Hence,

$$\mathrm{pOH} = 4.699$$

Given $$\mathrm{pH} + \mathrm{pOH} = 14$$, we find the $$\mathrm{pH}$$ as follows:

$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 4.699 = 9.301$$

Therefore, the $$\mathrm{pH}$$ of the resultant solution is 9.301.

The correct answer is:

Option C: 9.301

Ionic Equilibrum

Continue this paper

Similar Ionic Equilibrum PYQs

More practice from this question