Question

Calculate [NO3-] if 125 mL of 0.35 M NaNO3 is mixed with 450 mL of 1.1 M Mg(NO3)2. Please include some of your work as best as you can in the answer for full marks.

Answer 1

Answer: 1.76 M

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in L)}}` .....(1)

Molarity of
`NaNO_3` solution = 0.35 M

Volume of solution = 125 mL

Putting values in equation 1, we get:

a)
`0.35M=\frac{\text{Moles of}NaNO_3* 1000}{125ml}\\\\\text{Moles of }NaNO_3=(0.35mol/L* 125)/(1000)=0.044mol`

1 mole of
`NaNO_3` contains = 1 mol of
`NO_3^-`

Thus
`0.044mol` of
`NaNO_3` contain=
`(1)/(1)* 0.044=0.044` mol of
`NO_3^-`

b)
`1.1M=\frac{\text{Moles of}Mg(NO_3)_2* 1000}{450ml}\\\\\text{Moles of }Mg(NO_3)_2=(1.1mol/L* 450)/(1000)=0.495mol`

1 mole of
`Mg(NO_3)_2` contains = 2 mol of
`NO_3^-`

Thus
`0.495mol` of
`Mg(NO_3)_2` contain=
`(2)/(1)* 0.495=0.99` mol of
`NO_3^-`

Total
`[NO_3^-]=\frac {\text {total moles}}{\text {total volume}}=(0.044+0.99)/(0.575L)=1.76M`

Thus
`[NO_3^-` after mixing is 1.76 M