I have a 30% solution of Calcium Hydroxide. I want to make a 1% solution. How would I go about doing this? I tried using C1 x V1 = C2 x V2


closed as off-topic by Mithoron, Gaurang Tandon, airhuff, Tyberius, aventurin Jun 27 '18 at 4:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ How about you show your calculation as your approach seems to be right $\endgroup$ – Avnish Kabaj Jun 26 '18 at 19:09
  • 2
    $\begingroup$ You can't possibly have a 30% solution of calcium hydroxide. Other than that, your approach is right. $\endgroup$ – Ivan Neretin Jun 26 '18 at 19:18
  • $\begingroup$ right formula, wrong reality. I agree with Ivan. $\endgroup$ – MaxW Jun 26 '18 at 19:31
  • 1
    $\begingroup$ @user65526 Welcome to chemistry SE! If you haven't yet, take the tour and visit the help center. I would agree with Avnish, try to write up your attempt using Mathjax and then other users can fix any formatting issues to show you how it is done. $\endgroup$ – Tyberius Jun 26 '18 at 19:31

Let’s imagine you have 10 mL of this 30% $\ce{Ca(OH)2}$ solution. You are right to plug into the equation that you specified.

$$C_1*V_1 = C_2*V_2$$

$$30\%*10 mL = 1\%*V_2$$

$$\begin{split} V_2 & = \frac{30\%*10 mL}{1\%}\\ & = 300 mL \end{split}$$

This means that the final volume will be 300 mL. However, since there are already 10 mL of solution, you only need to add 300 mL - 10 mL = 290 mL of solution.

  • $\begingroup$ I'm glad you started off Let’s imagine.... Wikipedia list solubility as 1.73 g/L (20 °C). That's a long way from a 30% solution. // You're also wrong that you could add 290 ml water and 10 ml 30% Ca(OH)2 and get 300 ml of solution. You'd need to take 10 ml of base and dilute to 300 ml. $\endgroup$ – MaxW Jun 26 '18 at 20:40
  • $\begingroup$ @MaxW :) (characters needed) $\endgroup$ – Shady Puck Jun 26 '18 at 20:41
  • $\begingroup$ @MaxW Your comment is confusing me: What's the difference between 10 ml 30% Ca(OH)2 and 10 ml base? $\endgroup$ – Shady Puck Jun 26 '18 at 20:59
  • $\begingroup$ None. You'd of course need "about" 290 ml of water. But as an analytical chemist I was trying to point out that volumes aren't really additive. I greatly prefer to think of diluting the 10 ml to 300 ml total rather than adding 10 ml and 290 ml to get 300 ml. $\endgroup$ – MaxW Jun 26 '18 at 21:08
  • $\begingroup$ Ivan and Max W it is a 30% solution, which of course is mixed constantly. $\endgroup$ – user65526 Jun 27 '18 at 14:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.