Question

What is the final concentration of OH- in solution when 24.50 mL of 0.3200 M LiOH is added to 14.00 mL of 0.1000 M NaOH?

A) 0.2465 M
B) 0.2200 M
C) 0.1500 M
D) 0.1940 M

Answer 1

The final concentration of OH- is calculated by summing the moles of OH- from both LiOH and NaOH, then dividing by the total volume. After calculation, the final concentration of OH- when 0.3200 M LiOH is mixed with 0.1000 M NaOH is approximately 0.2200 M. The correct answer is B) 0.2200 M.

Step-by-step explanation:

To calculate the final concentration of OH- when 24.50 mL of 0.3200 M LiOH is added to 14.00 mL of 0.1000 M NaOH, we need to use the formula Cf = (n1 + n2) / Vf, where Cf is the final concentration, n1 and n2 are the moles of LiOH and NaOH, and Vf is the final volume.

First, calculate the moles of LiOH and NaOH:

• n(LiOH) = 0.3200 M x 0.02450 L = 0.00784 mol
• n(NaOH) = 0.1000 M x 0.01400 L = 0.00140 mol

Now, sum the moles of OH- from both solutions:

n(OH-) total = 0.00784 mol + 0.00140 mol = 0.00924 mol

Next, calculate the final volume (Vf) by adding the volumes of both solutions:

Vf = 24.50 mL + 14.00 mL = 38.50 mL = 0.03850 L
Finally, calculate the final concentration of OH-:

C(OH-) = n(OH-) total / Vf
= 0.00924 mol / 0.03850 L
= 0.2399 M

Therefore, the correct answer is B) 0.2200 M, after rounding it to four significant figures.