What volume of 0.1 M NaOH is required to neutralize 30 mL of 0.1 M HCl?
Well, let's do some math to figure it out. Remember, when you mix an acid with a base, they can neutralize each other. So, the number of moles of acid should be equal to the number of moles of base.
Given that the concentration of HCl is 0.1 M and the volume is 30 mL, we can find the number of moles of HCl by multiplying the concentration (0.1 M) by the volume (30 mL) and converting it to liters:
0.1 M x 0.03 L = 0.003 moles of HCl.
Since the acid and base will neutralize each other in a 1:1 ratio, we need the same number of moles of NaOH to neutralize the HCl. So, we want to find out how much NaOH is needed to have 0.003 moles.
To do that, we need to know the concentration of NaOH. You mentioned it's 0.1 M, so we can use that. Now, we rearrange the equation:
0.1 M x Volume (L) = 0.003 moles.
Solving for the volume, we can divide both sides by 0.1 M:
Volume (L) = 0.003 moles / 0.1 M.
Volume = 0.03 L or 30 mL.
So, you would need 30 mL of 0.1 M NaOH to neutralize 30 mL of 0.1 M HCl. Looks like a balanced equation!
To determine the volume of 0.1 M NaOH required to neutralize 30 mL of 0.1 M HCl, we will use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
The balanced equation tells us that the ratio of NaOH to HCl is 1:1, meaning that for every 1 mole of NaOH, 1 mole of HCl is neutralized.
Given that the concentration of both NaOH and HCl is 0.1 M, this means that 1 mole of NaOH is present in 1 L (1000 mL) of solution.
Now, let's calculate the number of moles of HCl in 30 mL of the 0.1 M HCl solution:
Moles of HCl = concentration (M) x volume (L)
= 0.1 M x 0.03 L
= 0.003 moles
Since the neutralization reaction is 1:1, we need an equal number of moles of NaOH to neutralize the HCl.
Therefore, we need 0.003 moles of NaOH.
To calculate the volume of 0.1 M NaOH required, we can use the equation:
Volume (L) of NaOH = moles of NaOH / concentration (M)
= 0.003 moles / 0.1 M
= 0.03 L
Converting this volume to milliliters:
Volume (mL) of NaOH = 0.03 L x 1000 mL/L
= 30 mL
Therefore, 30 mL of 0.1 M NaOH is required to neutralize 30 mL of 0.1 M HCl.
To determine the volume of 0.1 M NaOH needed to neutralize 30 mL of 0.1 M HCl, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
From the balanced equation, we can determine that the stoichiometric ratio between NaOH and HCl is 1:1. This means that for every 1 mole of HCl, we need 1 mole of NaOH for neutralization.
To find the number of moles of HCl in 30 mL of 0.1 M HCl, we can use the formula:
moles = concentration (M) × volume (L)
moles of HCl = 0.1 M × 0.030 L = 0.003 moles HCl
Since the stoichiometric ratio is 1:1, we need 0.003 moles of NaOH to neutralize the 0.003 moles of HCl.
Now, to calculate the volume of 0.1 M NaOH needed to obtain 0.003 moles, we can rearrange the formula:
volume (L) = moles / concentration (M)
volume of NaOH = 0.003 moles / 0.1 M = 0.03 L = 30 mL
Therefore, 30 mL of 0.1 M NaOH is required to neutralize 30 mL of 0.1 M HCl.
Calculate the number of moles for HCl in a 30mL sample of 0.1M HCl
30 x10^-3 L*(0.1M)=moles of HCl
The number of moles of HCl/the molarity of NaOH will give you the volume needed in L.
volume in L=moles of HCl/0.1 M NaOH
*** Multiply the volume by 10^3 if you need to report the volume in mLs.