Ionic Reactions and Equilibrium in Aqueous Solutions | Brief Solutions

B

Hydrochloric acid; $\mathrm{N_2H_5Cl}$, using the first protonation specified in the question.

Prepare a $0.010$ mol/L solution at $25^\circ\mathrm C$ and measure with a pH meter. A pH near $2.00$ indicates essentially complete ionization, supporting a strong acid.

$\mathrm{NaHCO_3\to Na^++HCO_3^-}$

$\mathrm{C_6H_5SO_3H+NaOH\to C_6H_5SO_3Na+H_2O}$

B, D

Higher pH promotes arsenic-acid deprotonation and increases negative charge on both arsenate species and the adsorbent. Stronger electrostatic repulsion reduces adsorption.

$\mathrm{2HCOOH+2CaSO_3\to Ca(HCOO)_2+Ca(HSO_3)_2}$

$\mathrm{CO_2+H_2O+CN^-\to HCO_3^-+HCN}$

$\mathrm{CO_3^{2-}+H_2S\rightleftharpoons HCO_3^-+HS^-}$;$K=10^3$

$\mathrm{SO_2+HCO_3^-\to HSO_3^-+CO_2\uparrow}$

$\mathrm{pH}\approx5.00$

A;$[\mathrm{Na^+}]=[\mathrm{H_2C_2O_4}]+[\mathrm{HC_2O_4^-}]+[\mathrm{C_2O_4^{2-}}]$

$\mathrm{pH}\approx2.71$

$\mathrm{pH}=10$