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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?

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    $\begingroup$ Unless we know what's in there, the answer can be anything. $\endgroup$ – Ivan Neretin Apr 24 at 20:27
  • $\begingroup$ 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. $\endgroup$ – Adrika De Apr 24 at 20:35
  • $\begingroup$ Good. Now what is the initial volume of the solution? $\endgroup$ – Ivan Neretin Apr 24 at 20:41
  • $\begingroup$ One litre , sir. $\endgroup$ – Adrika De Apr 24 at 20:47
  • $\begingroup$ 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? $\endgroup$ – Ivan Neretin Apr 24 at 20:52
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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

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