Question

A student is provided with a 0.15 M stock solution of hydrochloric acid (HCl). They take 2.00 mL of the solution and dilute it to 900.0 mL. What is the pH of the diluted solution?

Answer 1

We will start by determining the new concentration of the HCl solution. To do this we will calculate the dilution factor then use this dilution factor to calculate the new concentration:

`\begin{gathered} Dilutionfactor(D.F)=\frac{total\text{ }volume}{aliquot\text{ }volume}=(stock)/(working) \\ D.F=(900mL)/(2.00mL) \\ D.F=450 \\ \\ D.F=(0.15M)/(x) \\ 450=(0.15M)/(x) \\ x*450=0.15M \\ x=(0.15M)/(450) \\ x=3.33*10^(-4)M \end{gathered}`

Now we will calculate the pH,

`\begin{gathered} pH=-log[H^+] \\ pH=-log[3.33*10^(-4)] \\ pH=3.48 \end{gathered}`

Answer: The pH of the diluted solution 3.48,