A metal salt solution forms a yellow precipit | vedclass

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(B) $1$. The metal salt solution reacts with potassium chromate $(K_2CrO_4)$ in acetic acid to form a yellow precipitate. This indicates the presence

Vedclass · Accepted Answer

$Ba(NO_3)_2$

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(B) $1$. The metal salt solution reacts with potassium chromate $(K_2CrO_4)$ in acetic acid to form a yellow precipitate. This indicates the presence of $Ba^{2+}$ or $Pb^{2+}$ ions,as both form yellow chromates ($BaCrO_4$ and $PbCrO_4$).$2$. It forms a white precipitate with dilute $H_2SO_4$,which indicates the presence of $Ba^{2+}$,$Pb^{2+}$,or $Ca^{2+}$ (as $BaSO_4$,$PbSO_4$,and $CaSO_4$ are white).$3$. It gives no precipitate with $NaCl$. $Pb^{2+}$ forms a white precipitate of $PbCl_2$ with $NaCl$,whereas $Ba^{2+}$ does not form a precipitate with $NaCl$ because $BaCl_2$ is soluble.$4$. Therefore,the metal salt solution contains $Ba^{2+}$ ions,such as $Ba(NO_3)_2$.

$A$ metal salt solution forms a yellow precipitate with potassium chromate in acetic acid,a white precipitate with dilute $H_2SO_4$,but gives no precipitate with $NaCl$. The metal salt solution will consist of:

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