Question

Free residual chlorine concentration in water was measured to be 2 mg/l (as Cl\(_2\)). The pH of water is 8.5. By using the chemical equation given below, the HOCl concentration (in \(\mu\)moles/l) in water is ___________ (round off to one decimal place).
\[ {HOCl} \rightleftharpoons {H}^+ + {OCl}^-, \quad pK = 7.50 \] Atomic weight: Cl = 35.5
 

Hint:

Remember to convert units properly when dealing with mg/l and \(\mu\)moles/l. Also, use equilibrium constants and pH values to solve dissociation problems.

Answer 1

The equilibrium constant \(k\) for the dissociation of HOCl is given by: \[ k = \frac{[{HOCl}]}{[{OCl}^-][{H}^+]} \] Substituting the values: \[ 10^{7.5} = \frac{[{HOCl}]}{[{OCl}^-] \times 10^{-8.5}} \] Simplifying further: \[ 10^{-1} = \frac{[{HOCl}]}{[{OCl}^-]} \] This gives the relationship: \[ [{HOCl}] + [{OCl}^-] = \frac{2 { mg} \times 10^{-3}}{71} \] Converting to moles per liter: \[ [{HOCl}] + [{OCl}^-] = 2 \times 10^{-3} \, {moles/l} \times 10^6 = 2.56 \, \mu{moles/l} \] Thus, the concentration of HOCl in water is approximately \( \boxed{2.6} \, \mu{moles/l} \).