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Chemistry question ?
How many mL of 0.2423M NAOH are required to neutralize a mixture containing 10.00 mL of 0.2517M Nitric acid, 10.00 mL of 0.5159M carbonic acid, and 10.00 mL of 0.7129M phosphoric acid?
1 Answer
- Roger the MoleLv 75 days ago
HNO3 + NaOH → NaNO3 + H2O [balanced as written]
(10.00 mL) x (0.2517 M HNO3) x (1 mol NaOH / 1 mol HNO3) /
(0.2423 M NaOH) = 10.3879 mL NaOH for the nitric acid
2 NaOH + H2CO3 → Na2CO3 + 2 H2O
(10.00 mL) x (0.5159 M H2CO3) x (2 mol NaOH / 1 mol H2CO3) /
(0.2423 M NaOH) = 42.5836 mL NaOH for the carbonic acid
3 NaOH + H3PO4 → Na3PO4 + 3 H2O
(10.00 mL) x (0.7129 M H3PO4) x (3 mol NaOH / 1 mol H3PO4) /
(0.2423 M NaOH) = 88.2666 mL NaOH for the phosphoric acid
10.3879 mL + 42.5836 mL + 88.2666 mL = 141.2 mL NaOH total
