What volume of #12*mol*L^-1# #NaOH# is required to make a #1*L# volume of #4*mol*L^-1# concentration?

Question

Wyatt Adkins · Accepted Answer

We use the relationship #C_1V_1=C_2V_2# to get #V_1(NaOH)=1*L# The product #CxxV# gives us an answer in #"moles"#. And thus we need #3*Lxx4*mol*L^-1=12*mol# #NaOH#. And thus we take a #1*L# volume of #12*mol*L^-1# #NaOH#, and dilute this to #3*L#. So #"concentration"=(1*Lxx12*mol*L^-1)/(3*L)=4*mol*L^-1#.....as required.

Avery Adams · Answer

undefined To dilute a solution from a higher concentration to a lower concentration, you can use the formula:

$$C_1V_1 = C_2V_2$$

Where:
- $C_1$ = initial concentration (in mol/L)
- $V_1$ = initial volume (in L)
- $C_2$ = final concentration (in mol/L)
- $V_2$ = final volume (in L)

Given:
- $C_1$ = 12 mol/L
- $C_2$ = 4 mol/L
- $V_2$ = 1 L

Solve for $V_1$:

$$V_1 = \frac{{C_2 \times V_2}}{{C_1}}$$

$$V_1 = \frac{{4 \times 1}}{{12}}$$

$$V_1 = \frac{1}{3}$$

So, you would need $\frac{1}{3}$ L or $333.\bar{3}$ mL of the 12 mol/L NaOH solution to make a 1 L volume of 4 mol/L concentration.