${K_2}CrO_4$ and $Na_{2}CrO_{4}$ can be used to identify $Ba^{2+}$ and $Pb^{2+}$ both. If true, enter 1 and if false, enter 0.

Hint: cations are positively charged ions which are formed when atoms or groups of (covalently bonded) atoms lose the electrons. Cations possess fewer electrons than protons.
Detailed explanation: The presence of various cations like $Na^+$, $Pb^{2+}$, $Ba^{2+}$, $Al^{3+}$, $Mg^{2+}$ can be identified using different methods. As relevant to the question, the potassium chromate method has been discussed to identify $Ba^{2+}$ and $Pb^{2+}$ ions.

Complete step by step answer:
$Ba^{2+}$ and Pb 2+ ions can react with potassium chromate ${K_2}CrO_4$ to produce yellow precipitate of metal chromate ($BaCrO_4$ and $PbCrO_4$, respectively) as indicated below:
\[
  P{b^{2 + }} + {\text{ }}{K_2}Cr{O_4}{\text{ }} \to {\text{ }}PbCr{O_4}{\text{ }} \downarrow {\text{ }} + {\text{ }}2{K^ + } \\
  B{a^{2 + }} + {\text{ }}{K_2}Cr{O_4}{\text{ }} \to {\text{ }}BaCr{O_4}{\text{ }} \downarrow {\text{ }} + {\text{ }}2{K^ + } \\
\]

Additional information: Two other methods to detect the presence of cations are explained below:
Method 1:
1. Add few drops of dilute NaOH soln to a soln comprising of the unknown cations
2. White precipitate is formed indicating the presence of $Pb^{2+}$, $Ba^{2+}$, $Mg^{2+}$
3. Add excess of NaOH soln
4. If precipitate remains indicating the presence of $Mg^{2+}$ and $Ba^{2+}$. If precipitate disappears, it indicates the presence of $Al^{3+}$ or $Pb^{2+}$
5. Add dilute ${H_2}SO_4$ to a fresh sample of solution.
6. Disappearance of precipitate indicates the presence of $Mg^{2+}$. The appearance of white precipitate indicates the presence of $Ba^{2+}$.

Method 2:
$Ba^{2+}$ and $Pb^{2+}$ ions can react with sodium sulphate (${Na_2}SO_4$) to produce white precipitate of metal sulphate ($BaSO_4$ and $PbSO_4$, respectively) as indicated below:
\[
  P{b^{2 + }} + N{a_2}S{O_4}{\text{ }} \to {\text{ }}PbS{O_4} \downarrow + 2N{a^ + } \\
  B{a^{2 + }} + N{a_2}S{O_4}{\text{ }} \to {\text{ }} BaSO_4 \downarrow + 2N{a^ + } \\
\]
Note:
Barium chromate ($BaCrO_4$) is soluble in mineral acids, though only slightly soluble in acetic acid. On the other hand, an orange solution of barium dichromate is formed in strong acids.