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As many as hydroxide concentration:

[OH-] = 10-[14.0-pH] = 10-[14.0-11.2] = 10-2.8 =invlog(-2.8) = 1.6*10-3 mol/L

So, to neutralize you've to add 1.6*10-3 mole strong monoprotic acid (H+) to 1.0 L solution of pH 11.2

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Q: How much acid is required to neutralise pH level11.2?
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