The amount of water that will be created is 0.007 moles, while the amount of OH ions present is 0.123 M.
Given are V = 28 ml and molarity = 0.250 M for nitric acid.
Nitric acid has a molecular weight of 0.250 (28/1000) = 0.007 mol.
V = 53 ml and molarity = 0.320 M are provided for KOH.
KOH has a molecular weight of 0.320 (53/1000) = 0.01696 mol.
Nitric acid is the reaction's limiting reagent as a result.
One mol of nitric acid results in one mol of water, according to the chemical equation.
As a result, 0.007 mol of nitric acid will result in 0.007 mol of water.
Nitric acid is completely destroyed, but the remaining KOH will produce an excess of OH ions, which equals 0.01696 - 0.007 = 0.00996 moles.
The solution has a total volume of 28 + 53 = 81 ml, or 0.081 ml.
The amount of excess OH ions is equal to 0.00996/0.081 = 0.123 mol/L.
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