1. since the volume of sodium hydroxide is not specified, we will have to assume that the volume remains at a constant 1L after addition of NaOH. In this case, knowing that NaOH is a strong base and HNO3 is a strong acid, they will neutralize each other to form water.
H3O+(aq) + OH-(aq) --> 2H20(l)
0.48 moles HNO3 - .11 moles of NaOH = 0.37 moles H+ left over
Usually, you would be given the volume of NaOH added, since the [H3O+] is dependent on volume, such that:
.37 mol / (1L + V of NaOH added) = [HNO3+]
Since we are assuming the volume is constant at 1L, the [HNO3+] is equal to 0.37M after addition of 0.11 moles of NaOH.
2. The same concept applies; except that this time, the 0.48 moles of HNO3+ is completely neutralized by the moles of NaOH added, since the moles of NaOH are greater than 0.48.
Therefore, we are left with:
0.89 mol NaOH - 0.48 moles HNO3+ = .41 moles OH-
Once again assuming that the volume is held constant at 1L, the concentration of OH- is 0.41M. Since we need to calculate [H3O+] and this is a strong acid/strong base titration, we know that:
The water dissociation constant, Kw, is equal to 1.0x10^-14
Kw = [H3O+][OH-]
Since we have [OH-], we divide both sides by [OH-] to get [H3O+] by itself.
[H3O+] = 1.0x10^-14/0.41 = 2.4x10^-14M
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