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HCI + NaOH ->>

NaCl + H₂O
What volume of sodium hydroxide (NaOH) 0.9 M would be required to titrate 250 mL of hydrochloric acid (HCI)
0.25 M?
62.5 mL NaOH
(yellow)
69.44 mL NaOH
(purple)
90 mL NaOH
(blue)

Please help!!!!

User Platypus
by
7.4k points

1 Answer

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The balanced chemical equation for the reaction between hydrochloric acid (HCI) and sodium hydroxide (NaOH) is:

HCI + NaOH -> NaCl + H2O

From the equation, we can see that 1 mole of HCI reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.

First, let's calculate the number of moles of HCI in 250 mL of 0.25 M solution:

Molarity (M) = moles of solute / volume of solution (L)

0.25 M = moles of HCI / 0.25 L

moles of HCI = 0.25 L x 0.25 M = 0.0625 moles

Since 1 mole of NaOH reacts with 1 mole of HCI, we will need 0.0625 moles of NaOH to neutralize the HCI.

Now, let's calculate the volume of 0.9 M NaOH solution needed to provide 0.0625 moles of NaOH:

Molarity (M) = moles of solute / volume of solution (L)

0.9 M = 0.0625 moles of NaOH / volume of NaOH solution (L)

volume of NaOH solution (L) = 0.0625 moles / 0.9 M = 0.0694 L = 69.44 mL

Therefore, 69.44 mL of 0.9 M NaOH solution would be required to titrate 250 mL of 0.25 M HCI solution.

User Cletus
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8.7k points