Question

The cation content of a water sample is determined by exchanging the cations for protons on strong acid and ion-exchange resin. An 25 mL sample was added to a 2 g resin. After reaching equilibrium the resin is filtered off and the solution titrated with alkali. 18 mL of 0.020 M NaOH are required to reach the endpoint. Calculate the normal concentration of cations ( the number of equivalents present in 1 L of sample).Your answer must be in order of 10^(-13) and in 2 d.p.

Answer 1

R-H (resin) + M⁺ (cation) $\rightarrow$ R-M +H ⁺

H⁺ + OH^-$\rightarrow$ H₂O

[ 1 molar of NaOH = equivalent NaOH]

Number of gram equivalent of NaOH = Number of gram equivalent of H⁺ present in the solution.

and Number of gram equivalent of H⁺ present = gram equivalent of cation .

Now, moles of NaOH = gram equivalent of NaOH = volume(L)× molarity

Given volume =18 ml = 18×10^-3 L

Then, moles of NaOH = 18×10^-3 × 0.020 = 3.6×10^-4 .

So, gram equivalent of cation = 3.6×10^-4

Or, normality (the number of equivalentd per 1L) of cations =

[(Gram equivalent ×1000) ÷ volume of sample solution (mL)]

= [(3.6×10^-4×1000)÷25]

= 0.0144

= 1.44×10^-2 (gram- equivalent/L)