# Calculation of pH in acid base reaction

This problem(from ionic equiibrium) states:

We are given $$\pu{40 mL}$$ of $$\pu{0.1 M}$$ $$\ce{HCl}$$ and $$\pu{10 mL}$$ of $$\pu{0.45 M}$$ $$\ce{NaOH}$$ in two separate beakers. These two solutions are mixed. Find the $$\mathrm {pH}$$ of the resultant solution formed. Consider the concentration of the ions formed by self dissociation of water to be negligible.

### My method

Number of moles of $$\ce{H+}$$ ion in the first solution is $$\pu{0.004 mol}$$ and the number of moles of $$\ce{OH-}$$ in the second solution is $$\pu{0.0045 mol}$$.

Now, after mixing the two solutions, the resulting concentration of $$\ce{H+}$$ ion becomes $$\pu{0.08 M}$$ and that of the $$\ce{OH-}$$ will become $$\pu{0.09 M}$$. Now these $$\ce{H+}$$ and $$\ce{OH-}$$ ions will react with each other such that the equilibrium constant of their reaction becomes $$10^{-14}$$.

So I formed the equation

$$(0.08-x)(0.09-x) = 10^{-14}$$

From here we can calculate $$x$$ and subsequently we can get the concentration of $$\ce{H+}$$ ion in the final solution.

I want to know is this the correct way to solve the problem?

The answer to this problem is given as $$12$$, which I doubt can be obtained from my way of doing the problem.

• While this may be in the section on "ionic equilibrium," this is simply a limiting/excess reagents problem. – Zhe Jul 20 at 13:32
• @Zhe can you please help me with an answer on how to solve this problem and why is my method incorrect. – It's probable Jul 20 at 13:49
• Instead of solving it using Kw try using the principle of neutralisation $\ce{H+ + OH- -> H2O}$ – Safdar Jul 20 at 13:52
• @Safdar Ok let me try, but why is it wrong if I solve using Kw? – It's probable Jul 20 at 13:59
• Hint: There are 0.5 millimoles of $\ce{OH-}$ in the final solution. Can you calculate the $\mathrm{pH}$ now? – Aniruddha Deb Jul 20 at 15:04

As stated by Zhe, it is difficult to solve the equation as the value of x would be very close to $$0.08$$. However there is a workaround, if you wish to solve it this way using $$K_\mathrm{w}$$ from the side of the $$\ce{H+}$$ ion. The equation that you have is: $$\tag{1} (0.08-x)(0.09-x)=10^{-14}$$

Instead of solving for $$x$$ here, we rather say $$Y = 0.08-x$$. Here $$Y$$ acts as the final concentration of $$\ce{H+}$$ in the solution (not counting dissociation of water). Using this, we get equation ($$1$$) to become:

$$Y(0.01+Y)=10^{-14}$$

Now, we assume that since the product is a very small number, $$Y + 0.01 \approx 0.01$$. Now equation ($$1$$) becomes: $$0.01Y=10^{-14}$$

Therefore, $$Y = 10^{-12}$$, pH is equal to $$12$$.

No, you should not do the problem your way.

Given

$$(0.08−x)(0.09−x)=10^{−14}$$

You can solve for $$x$$ here, but its value will be quite close to $$0.08$$. This is the only way to make the product a small number, by making the one of the multiplicands very small.

However, $$x$$ will be so close to 0.08 that will have trouble figuring out the value of $$0.08 - x$$, which is what you want. Therefore, the right answer is obscured by your method of solving the problem.

In this problem, you can work around that by calculating the pH value from the hydroxide ion concentration, which, by the argument above, will be roughly $$0.01\ \mathrm{M}$$, consistent with a pH of 12. However, in the most general case, you may not be able to do that to determine what the final equilibrium concentrations are.

• (+1) Very nice job explaining to the OP why their method is not optimal! ;) – Ed V Jul 20 at 16:24