Question

In sample D we poured the 50 cm3 hot water. Chlorine gas with hot water react like this:
2Cl2(g)+2H2O(l) ---> 4HCl(aq)+O2(g)
Please find in sample D formed hydrochloric acid's solutions pH if it's known that 8,85 cm3 oxygen gas been released. Gas volume measured by normal conditions. Don't worry about volume's difference during reaction. Please give a consistent answer.

Answer 1

Moles of oxygen released in $8.85 \space cm^3(8.85 \space ml)=\frac{8.85}{22400}\space moles=0.0004 \space moles.$

$2Cl_2(g)+2H_2O(l) ---> 4HCl(aq)+O_2(g)$

As per stoichiometry of the reaction ,4 moles of $HCl$ are released when 1 mole of $O_2$ is formed

So,moles of $HCl$ .released=$4 \times 0.04 \times 10^{-2}=0.16 \times 10^{-2} \space moles$

Volume of solution=$50 \space cm^3=50 \space ml=50 \times 10^{-3} ltr$

Concentration of $HCl \space or \space H^+ =\frac{0.16 \times 10^{-2}}{50 \times 10^{-3}}=3.2 \times 10^{-2} M$

pH of solution $=-log[H^+]=-log[3.2 \times 10^{-2}]=1.495$