how many mL of 0.15 M NaOH solution are required to neutralize 35.00 mL of 0.22 M HCl
mL NaOH x M NaOH = mL HCl x M HCl
The easy way is simple substitution.
mL NaOH x M NaOH = mL HCl x M HCl
mL NaOH x 0.15M = 35.00mL x 0.22M
Solve for mL NaOH. I get (35.00*0.22/0.15) = approximately 51 mL NaOH.
The only problem with working all titration problems that way is that not all titrations are 1:1 as these are; for example,
NaOH + HCl ==> NaCl + H2O
The more fundamental way, but a little long, and it works with all types of titrations is to convert to moles.
1. Write the equation and balance it.
NaOH + HCl => NaCl + H2P
2. How many moles HCl do we have? That is M x L = 0.22M x 35.00 mL = 7.70 moles.
3. Usint the coefficients in the balanced equation, convert moles HCl to moles NaOH.
7.70moles HCl x (1 mole NaOH/1 mole HCl) = 7.70 x (1/1) = 7.70 moles NaOH.
4. Then the definition of molarity is M = moles/L. 0.15M = 7.70moles/L and
L = 7.70/0.15 = approximately 51.
Thanks alot! So Ma x Va= Mb x Vb or Ma=MbVb/Va :)
To find the volume of the NaOH solution required to neutralize the HCl solution, you can use the concept of stoichiometry and the balanced chemical equation for the reaction between NaOH and HCl:
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
From the balanced equation, you can see that 1 mole of NaOH reacts with 1 mole of HCl to form 1 mole of NaCl. Therefore, the mole ratio of NaOH to HCl is 1:1.
To calculate the amount of NaOH in moles, you can use the equation:
Moles = Concentration x Volume (in liters)
For HCl:
Moles of HCl = 0.22 M x 0.035 L = 0.0077 moles of HCl
Since the mole ratio is 1:1, the moles of NaOH required will also be 0.0077 moles.
Now, to find the volume of the NaOH solution, you can rearrange the equation above:
Volume (in liters) = Moles / Concentration
For NaOH:
Volume of NaOH = 0.0077 moles / 0.15 M ≈ 0.0513 L
To convert the volume to milliliters (mL), you simply multiply by 1000:
Volume of NaOH = 0.0513 L x 1000 = 51.3 mL
Therefore, approximately 51.3 mL of the 0.15 M NaOH solution is required to neutralize 35.00 mL of the 0.22 M HCl solution.