What is the pH of 8 x 10^-8 M HCl? Solution This is a solution of HCl. The solut

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This is a solution of HCl. The solution is acidic. The ph must therefore be less than 7.00 If you take HCl to dissociate completely , then [H+] = 8*10^-8M If you apply the formula: pH = -log [H+] pH = -log 8*10^-8 pH = 7.097 This answer must be wrong because a solution of HCl cannot have pH > 7.00 What must be taken into account is the [H+ ] from the auto-ionisation of water. At normal conditions this is [H+] = 1*10^-7 You will not solve this problem only by using the quadratic equation in your calculations. Unless you take the [H+] from the water into account you will always get a result above 7.00. The full method of working this out is rather long and tedious. The quick "rough and ready "way is to add the [H+] from the acid to the [H+] from the water , and apply the formula: Total [H+] = (1*10^-7) + (8*10^-8) = 1.8*10^-7 Now apply your formula: pH = -log ( 1.8*10^-7) pH = 6.74. This is not exactly technically correct, but the result is at least below 7.0 and it is a good approximation of the correct answer. A good site that explains this fully is ChemTeam. Check it out and see more about this. I am pleased to see that I am in good company of ChemTeam in using the "quick and easy"method above for this problem. But they do give you the full complicated method , which you might want to read up. As you see it is not a matter of using the quadratic method - it is understanding where the total [H+] comes from

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