We have 1 Na on the left side and 3 Na on the right side in the first equation. So, we put a 3 in front of NaOH on the left side. The equation becomes:
\[3\mathrm{NaOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{Na}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2}
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