Calculating the pH of an Acid-Water Mixture

[Gregory.gags]

A flask contains 500 mL of an acid with a pH of 4. If 500 mL of water is added, what will the pH of the new solution be?

I don't want to know the answer please*

I'm just wondering if it is as easy as adding the two pH's ie. 7 + 4

or do I have to work out the H⁺ of each ie. 10^-7 and 10^-4 and then add, multiply or divide those. And why?

 

[Nessdude14]

pH is defined by the following equation:
\large pH = -log[H^{+}]
It's not quite as simple as adding the pHs, but when you halve the concentration of H⁺ (by doubling the volume of the solution), the new pH (which we'll call pH₂) is:
\large pH_{2} = -log\frac{[H^{+}]}{2}
Which, by laws of logarithms, is equivalent to:
\large pH_{2} = -log([H^{+}])-log(2)
Since log(2)=-0.301 (approximately), we can say:
\large pH_{2} ≈ pH_{1}+0.3
where pH₁ is the pH before the water was added.

 

[AGNuke]

pH is not a linear function you can go on adding, it is a logarithmic function. Its addition depends totally upon addition of [H⁺]

First, we have to calculate the initial concentration of H⁺. It is 10^-4. Now, we diluted it with equal amount of water, so the concentration becomes half.

Note that water also consist [H⁺] = 10^-7, but it can be neglected in front of [H⁺] of initial solution.

Now pH = -log₁₀[H⁺]; [H⁺] = 5 x 10^-5

pH comes out to be 4.3010

 

[Gregory.gags]

pH is defined by the following equation:
\large pH = -log[H^{+}]
It's not quite as simple as adding the pHs, but when you halve the concentration of H⁺ (by doubling the volume of the solution), the new pH (which we'll call pH₂) is:
\large pH_{2} = -log\frac{[H^{+}]}{2}
Which, by laws of logarithms, is equivalent to:
\large pH_{2} = -log([H^{+}])-log(2)
Since log(2)=-0.301 (approximately), we can say:
\large pH_{2} ≈ pH_{1}+0.3
where pH₁ is the pH before the water was added.

ok I got pH₂=3.699 after substituting 10^-4 for [H⁺] after working through that. This makes sense since the solution is diluted the pH would be less, no?

 

[AGNuke]

No No No... Diluting Acidic solution will increase the pH. See, pH is MINUS Log of concentration of H⁺. If we dilute the solution, the concentration decreases, its log decreases, but pH increases. (see negative sign).

Here we halved its concentration... So the pH increases. By calculating the -log of 10^-4/2; we get 4.3010, not 3.699.

3.699 is the pH of the solution if you actually concentrate it. (eg. like boil the solution until half the initial volume remains).

Remember; Dilution increases pH in acidic solutions, decreases pH in basic solution.

 

[Nessdude14]

ok I got pH₂=3.699 after substituting 10^-4 for [H⁺] after working through that. This makes sense since the solution is diluted the pH would be less, no?

With an H⁺ concentration of 10^-4, your pH₁ (before addition of water) would be 4. When you add the water, you add 0.3 to the pH, making the new pH 4.3. The diluted pH is higher because water is neutral with a pH of 7, so unless the water causes a reaction, adding water will always bring the pH closer to 7.

I should mention that the equations I made will only work for an acidic solution. With a neutral or basic solution, the addition of water does not halve the H⁺ concentration as we assumed in the equations. The assumption we made is accurate to 2 significant figures for any pH below 6, and it's accurate to 4 sig figs below pH 5.