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'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?

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 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.
 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

 Quote by 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.