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








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 3 years ago.








[Improve this question](/posts/132459/edit) 



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?




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


 3