Question

How much NaOH (in grams) is needed to prepare 463 mL of solution with a pH of 10.020?

Answer 1

Answer:
`1.94* 10^(-3)g` of NaOH

Step-by-step explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

`pH=-\log [H^+]`

Putting in the values:

`10.020=-\log[H^+]`

`[H^+]=9.55* 10^(-11)`

`[H^+][OH^-]=10^(-14)`

`[OH^-]=(10^(-14))/(9.55* 10^(-11))=1.05* 10^(-4)M`

`NaOH\rightarrow Na^++OH^-`

`Molarity=\frac{moles* 1000}{\text {Volume in ml}}`

`1.05* 10^(-4)M=(moles* 1000)/(463ml)`

moles =
`4.86* 10^(-5)`

Mass of
`NaOH=moles* {\text {Molar mass}}=4.86* 10^(-5)* 40=1.94* 10^(-3)g`

Thus
`1.94* 10^(-3)g` of NaOH is needed to prepare 463 mL of solution with a pH of 10.020