NH 3 + H 2 O → NH 4 + + OH−
C.
NO 2 − + H 2 O → HNO 2 + OH−
D.
4 . 0.067
At the equivalence point,Plugging into the formula,5 . 10.2
First, add enough HCl to neutralize the solution. Since both the acid and the base are 1 M, 10
mL of HCl will neutralize 10 mL of NaOH. This produces 20 mL of 0.5 M NaCl solution.Next, calculate how much more HCl must be added to produce a [H+] of 1 × 10 −2. Let x be the
amount of HCl to be added. The total volume of the solution will be (20 + x) mL. Since this is
now a dilution problem, the amount of HCl to be added can be found by using the formula:When this equation is solved, x is found to have the value of 0.2. The final volume is 20.2 mL, so
10.2 mL of HCl was added to the original NaOH solution.6 . (H 2 O, OH−) and (NH 4 +, NH 3 )
NH 4 + is the conjugate acid of the weak base, NH 3 ; OH− is the conjugate base of the weak acid,
H 2 O. The reaction in question is: