What quantity of 2M HCl is needed to bring 1M water (pH 7.0) down to pH 5.5?
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Answer:
If we assume V = 1 L: pH = -log[H+] = 5.5 [H+] = 10^-5.5 = 3.16 * 10^-6 M [HCl] = 2 M, so there are 2 mol of H+ in 1 L of HCl We need 3.16*10^-6 mol H+, so we should add 1.58*10^-6 L of HCl for every liter of water.
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Other answers
If we assume V = 1 L: pH = -log[H+] = 5.5 [H+] = 10^-5.5 = 3.16 * 10^-6 M [HCl] = 2 M, so there are 2 mol of H+ in 1 L of HCl We need 3.16*10^-6 mol H+, so we should add 1.58*10^-6 L of HCl for every liter of water.
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