What volume of .200M NaOH will be required to react completely with 32 mL of .15M HNO3
mL x M = mL x M
Convert mL to L
(32ml)(1L/1000mL) = 0.032L
(0.032L)(0.15M HNO3/1L) = 0.0048 moles HNO3
0.0048 mol = (.200M NaOH)(? L NaOH)
0.0048 / .200 = 0.024 L NaOH
0.024L (1000mL/1L) = 24ml NaOH
Good Luck with the rest of your Wiley Homework and hope this helps! :)
Oh, and by the way, you didn't post the equation for this question, so my answer above is assuming that there is an equal amount moles for NaOH and HNO3.
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To determine the volume of NaOH required to react completely with HNO3, we need to use the balanced chemical equation between NaOH and HNO3.
The balanced chemical equation is:
NaOH + HNO3 -> NaNO3 + H2O
From the balanced equation, we can see that the stoichiometric ratio between NaOH and HNO3 is 1:1. This means that 1 mole of NaOH reacts with 1 mole of HNO3.
First, let's calculate the number of moles of HNO3 using the given concentration and volume of HNO3:
Moles of HNO3 = concentration × volume
Moles of HNO3 = 0.15 M × 0.032 L
Moles of HNO3 = 0.0048 mol
Since the stoichiometric ratio between NaOH and HNO3 is 1:1, we need the same number of moles of NaOH to react completely with HNO3.
Now, let's find the volume of NaOH needed using the moles of NaOH and its concentration:
Moles of NaOH = 0.0048 mol
Concentration of NaOH = 0.200 M
Volume of NaOH = Moles of NaOH / Concentration of NaOH
Volume of NaOH = 0.0048 mol / 0.200 M
Volume of NaOH = 0.024 L or 24 mL
Therefore, 24 mL of 0.200 M NaOH will be required to react completely with 32 mL of 0.15 M HNO3.