Question

Find the milliequivalents of${\mathit{H}}^{\mathbf{+}}$released if100mLof$\mathbf{26}\mathbf{.}\mathbf{3}\mathit{m}\mathit{M}\mathit{N}{\mathit{i}}^{\mathbf{2}\mathbf{+}}$were loaded on a cation exchange column in the${\mathit{H}}^{\mathbf{+}}$form.

Short answer

The milli equivalents of $H^{+}$ that will be replaced by 100mL is$26.3mMN{i}^{2+}is5.26meq$

Step-by-step solution

Concept used

Given that the volume and concentration of $N{i}^{2+}is100mL$and $N{i}^{2+}is26.3mM$charge is +2

In the cation exchange column, calculate the milli equivalents of which will be replaced by 100mL of $26.3mMN{i}^{2+}$

Determine the meq of H+ released

Consider that the volume and concentration $N{i}^{2+}of100mL$and $N{i}^{2+}is26.3mM$charge is +2 .

$$\begin{gathered} eq=n\times \left(molesof{M}^{n+}\right) \\ =2\times 0.100L\times 0.0263mol/L \\ =5.26meq \end{gathered}$$

The supplied concentration, volume, and charge are put into the above equation to produce milli equivalents of $H^{+}$ which will be replaced in the cation exchange column by 100mL of 26.3mM

In the cation exchange column, the milli equivalents of H⁺ that will be replaced by 100mL of $26.3mMN{i}^{2+}$are $5.26meq$