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(adsbygoogle = window.adsbygoogle || []).push({});

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# PH calculation

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