calculate the volume of 0.0321M NaOH that will be required to neutralize 25.00 mL of a 0.0399M hydrochloric acid solution.
M1v1=m2v2
since were trying to find out the volume I would change the equation to m2 times V2 divided by M1
0.0399 X 0.025/0.0321 =0.031L
I don't know if this is right
It looks ok to me.
To calculate the volume of 0.0321M NaOH required to neutralize 25.00 mL of a 0.0399M HCl solution, you can use the equation:
M1V1 = M2V2
Where:
M1 = concentration of the acid (HCl) in mol/L
V1 = volume of acid (HCl) solution in L
M2 = concentration of the base (NaOH) in mol/L
V2 = volume of base (NaOH) solution in L
In this case, the concentration of HCl (acid) is 0.0399M and the volume is 25.00 mL (which needs to be converted to L). The concentration of NaOH (base) is 0.0321M, and we need to find the volume.
Let's solve the equation step by step:
Step 1: Convert the volume of the HCl solution from mL to L:
V1 = 25.00 mL = 25.00 mL * (1 L / 1000 mL) = 0.025 L
Step 2: Substitute the values into the equation:
(0.0399 M) * (0.025 L) = (0.0321 M) * V2
Step 3: Solve for V2:
V2 = (0.0399 M * 0.025 L) / 0.0321 M
Now let's calculate:
V2 = 0.0009975 L
The volume of 0.0321M NaOH required to neutralize 25.00 mL of a 0.0399M HCl solution is approximately 0.0009975 L or 0.9975 mL.