5 mL O.1 M HCL...How much 0.1 M NaOH is needed to neutralize the acid? Thank you
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5 mL?
5*0.1 = 0.1*x
solve for x.
To find out how much 0.1 M NaOH is needed to neutralize the 5 mL 0.1 M HCl, you need to use the concept of stoichiometry and the balanced chemical equation representing the neutralization reaction.
First, let's write the balanced equation for the neutralization reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
From the balanced equation, you can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.
Since the concentration of both HCl and NaOH is 0.1 M, it means that 1 L of the solution contains 0.1 moles (0.1 M = 0.1 moles/L).
Now, we can calculate the amount of moles of HCl present in 5 mL (0.005 L) using the formula:
moles of HCl = concentration (M) x volume (L)
= 0.1 moles/L x 0.005 L
= 0.0005 moles
Since the stoichiometry of the reaction is 1:1, it means that the reaction requires the same amount of moles of NaOH to neutralize the HCl.
Therefore, you would need 0.0005 moles of NaOH to neutralize the 5 mL of 0.1 M HCl.
To find the volume of the 0.1 M NaOH solution required, you can rearrange the formula and solve for volume:
volume (L) = moles of NaOH / concentration (M)
= 0.0005 moles / 0.1 moles/L
= 0.005 L or 5 mL
Thus, you would need 5 mL of 0.1 M NaOH to completely neutralize the 5 mL of 0.1 M HCl.