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Chemistry
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1,NaOH(s) was added to 1.0 L of HNO3(aq) 0.48 M. 

i,Calculate [H3O+] in the solution after the addition of 0.11 mol of NaOH(s)

ii,Calculate [H3O+] in the solution after the addition of 0.89 mol of NaOH(s).

Apr 2nd, 2015

The reaction equation is NaOH (s) + HNO3 (aq) => NaNO3 (aq) + H2O (l)

The original amount of HNO3 is 1.0 L * 0.48 mol/L = 0.48 mol. Since the coefficients in the reaction are equal for NaOH and HNO3, the amount of acid neutralized is equal 0.11 mol in the first case. The remaining amount of acid will be 0.37 mol. Since HNO3 is a strong acid, the concentration of hydronium will be 0.37 M, too.

pH = - log(0.37) = 0.43.

In the second case all nitric acid will be neutralized and there will be 0.89 - 0.48 = 0.39 mol of NaOH left.

Sodium hydroxide is a strong base, so the concentration of [OH-] will be also 0.39.

pOH = - log(0.39) = 0.41 and pH = 14 - pOH = 14 - 0.41 = 13.59

Apr 2nd, 2015

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