Question

Which of the following compounds is neutral to litmus?

• A. ${NaNO}_3$
• B. ${CuSO}_4.5H_{2}O$
• C. ${NaHCO}_3$
• D. $Ca{\left(OH\right)}_2$

Answer 1

The correct option is A

${\mathrm{NaNO}}_3$

The explanation for the correct option (A):

• Salt is a chemical compound that is produced by replacing the Hydrogen ($\mathrm{H}$) in acid with a metal.
• For example, Hydrochloric acid ($\mathrm{HCl}$) is an acid. If $\mathrm{H}$ of $\mathrm{HCl}$ is replaced by a metal, Sodium ($\mathrm{Na}$), then a Sodium chloride ($\mathrm{NaCl}$) salt is formed.
• The pH of salt solutions can be determined by the hydrolysis of salts.
• The salts of strong bases and strong acids produce neutral solutions.
• The salts of strong acids and weak bases produce acidic solutions.
• The salts of strong bases and weak acids produce basic solutions.
• Sodium nitrate (${\mathrm{NaNO}}_3$) is a salt of a strong base, Sodium hydroxide ($\mathrm{NaOH}$) and a strong acid, nitric acid (${\mathrm{HNO}}_3$).
• ${\mathrm{NaNO}}_3\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{HNO}}_3\left(\mathrm{aq}\right)$
• Hence, ${\mathrm{NaNO}}_3$ forms a neutral solution.
• Litmus turns blue in basic solution and it turns red in acidic solution. It remains neutral for a neutral solution.
• Hence, ${\mathrm{NaNO}}_3$ is neutral to litmus.

The explanation for the incorrect options:

(B)

• Copper sulfate pentahydrate (${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$) is a salt.
• The hydrolysis of Copper sulfate (${\mathrm{CuSO}}_4$) produces Copper hydroxide ($\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$) and Sulfuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$).
• ${\mathrm{CuSO}}_4\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{Cu}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)$
• Copper hydroxide is a relatively weaker base compared to Sulfuric acid, which is a strong acid. Thus, ${\mathrm{CuSO}}_4$ is a salt of strong acid and weak base.
• $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$
• ${\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$
• Hence, the solution of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ is acidic.

(C)

• Sodium hydrogen carbonate (${\mathrm{NaHCO}}_3$) is a salt of a strong base $\mathrm{NaOH}$ and a weak acid, Carbonic acid (${\mathrm{H}}_2{\mathrm{CO}}_3$).
• ${\mathrm{NaHCO}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)$
• Hence, the solution of ${\mathrm{NaHCO}}_3$ is basic.

(D)

• Calcium hydroxide ($\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$) is a strong base that ionizes completely to produce Hydroxide ions (${\mathrm{OH}}^{-}$) and Calcium ions (${\mathrm{Ca}}^{2+}$).
• $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)\to {\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Final Answer: Hence, option (A) is correct. ${\mathrm{NaNO}}_3$ is neutral to litmus.