# How to calculate the amount of water required to increase the pH of a solution from 2.1 to 4? [closed]

After working out the problem with certain approximation of considering the initial pH=2, my answer came out to be around 1 litre of water. But the correct answer is 79 litres of water. Can anyone kindly help me with this problem and if possible tell me what I probably missed out on?

• Unless we know what's in there, the answer can be anything. – Ivan Neretin Apr 24 at 20:27
• Exactly sir, I don't understand the same. But I think we can assume it to be a litre of strong acid in the initial case since the question came under calculation of pH of strong acids. – Adrika De Apr 24 at 20:35
• Good. Now what is the initial volume of the solution? – Ivan Neretin Apr 24 at 20:41
• One litre , sir. – Adrika De Apr 24 at 20:47
• Then the correct answer makes sense, and your one doesn't. To tell more, I need to know more. How did you come to that answer of yours? – Ivan Neretin Apr 24 at 20:52

If the acid is $$\ce{HCl}$$, the problem is how to dilute HCl with pH $$2.1$$ up to pH $$4$$. But pH $$2.1$$ means that $$\ce{[H^+] = 10^{-2.1} = 7.94 10^{-3} M = c_1}$$. Applying the expression $$\ce{c_1V_1 = c_2V_2}$$, with $$\ce{V_1 = 1 }$$ L, and $$\ce{c_2 = 10^{-4}}$$ M, the final volume $$\ce{V_2}$$ will be $$79.4$$ liters. It also means that the volume of water to be added to $$1$$ liter is $$79.4 - 1 = 78.4$$ liters