Question

Calculate the amount of KOH required neutralizing 15 milliequivalents of c.
(A) 0.84 g
(B) 0.80g
(C) 0.89 g
(D) none of these

Answer 1

Hint: A neutralization reaction is when an acid and a base react to form water and a salt involves the combination of ${{H}^{+}}$ and $O{{H}^{-}}$ ions to generate water. The neutralization of a strong acid and strong base has pH equal to 7. The neutralization of a strong acid and weak base will have a pH of less than 7, consequently the resulting pH when a strong base neutralizes a weak acid will be greater than 7.

Complete answer:
A neutralization reaction occurs when a strong acid reacts with a strong base to produce water and salt.
In this case, ${{N}_{2}}{{O}_{5}}$ (acid) will react with sodium hydroxide (KOH) to form water and potassium nitrate.
     \[{{N}_{2}}{{O}_{5}}+KOH\to KN{{O}_{3}}+{{H}_{2}}O\]
15meq ${{N}_{2}}{{O}_{5}}$ will require 15 milliequivalents of KOH for neutralization.
The molar mass of KOH =56g/mol
Hence, 15 milliequivalents of KOH corresponds to,
The amount of KOH = molar mass of KOH X moles of KOH
               = 15 X 56 = 840mg = 0.84 g of KOH
Hence, the amount of KOH required to neutralize, 15 milliequivalents of ${{N}_{2}}{{O}_{5}}$ =0.84g

The correct answer is option (A) = 0.84 g

Note:
The oxides that one uses to form acids and bases in aqueous solution often have reactivity that reflects their acidic or basic character. For example, \[L{{i}_{2}}O,CaO\And BaO\] react with water to form basic solutions and can react with acids directly to form salts. Likewise, ${{N}_{2}}{{O}_{5}},C{{O}_{2}}\And S{{O}_{3}}$ form acidic aqueous solutions and can react directly with bases to give salts.