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a 25.0-ml sample of h3po4 is titrated with naoh. if 29.2 ml of 0.738 m naoh is needed to reach the endpoint, what is the concentration (m) of the h3po4 solution?

User Vakata
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Hi! To find the concentration (M) of the H3PO4 solution, we will use the titration process where the H3PO4 is titrated with NaOH until the endpoint is reached. We are given a 25.0 mL sample of H3PO4 and 29.2 mL of 0.738 M NaOH is needed to reach the endpoint.

Step 1: Write the balanced chemical equation.
H3PO4 + 3NaOH → Na3PO4 + 3H2O

Step 2: Calculate moles of NaOH used in the reaction.
moles of NaOH = volume (L) × concentration (M)
moles of NaOH = (29.2 mL × (1 L / 1000 mL)) × 0.738 M
moles of NaOH = 0.0215 mol

Step 3: Use the stoichiometry from the balanced equation to find moles of H3PO4.
1 mol H3PO4 reacts with 3 mol NaOH
moles of H3PO4 = moles of NaOH ÷ 3
moles of H3PO4 = 0.0215 mol ÷ 3
moles of H3PO4 = 0.00717 mol

Step 4: Calculate the concentration of H3PO4.
concentration (M) = moles of H3PO4 ÷ volume of H3PO4 (L)
concentration (M) = 0.00717 mol ÷ (25.0 mL × (1 L / 1000 mL))
concentration (M) = 0.287 M

So, the concentration of the H3PO4 solution is 0.287 M.

User JoseLinares
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