Dilution of 0.01M HCN Calculating Volume of Water to Add

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The discussion revolves around diluting a 0.01M solution of HCN to achieve a pH of 6. The necessary concentration of HCN for this pH, calculated using the acid dissociation constant (Ka) of HCN, is determined to be 2.08*10^-3M. To find the new total volume required for this concentration, the original volume of 40cm3 is multiplied by the ratio of the initial concentration to the desired concentration, resulting in a new volume of 192cm3. Consequently, the volume of water needed to be added for the dilution is calculated to be 152cm3. The calculations have been verified by another participant in the discussion.
josephcollins
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I have a short question concerning the dilution of a 0.01M solution of HCN. I need to dilute 40cm3 of the above to give a solution with a pH of 6. Given the Ka of HCN is 4.8*10^-10 I obtain the concentration of HCN necessary in the new volume:2.08*10^-3M. Therefore the volume has to be multiplied by 0.01/2.08*10^-3 to give a new volume of 192cm3. Is it therefore that the volume of water necessary to add is 192-40=152cm3? Thx, Joe
 
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Yes, you are correct. I have checked it using BATE :wink:
 
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