What is the pH of the solution that results from adding 40mL of .2M NaCl to 60mL of .8M HCl?

Answer 1

Find the moles of HCl:

n = [HCl]*V2 = 0.8*0.06 = 0.048 mol.

Find the final volume:

V = V1 + V2 = 0.04 + 0.06 = 0.1 L

Find the Molarity of HCl in the final solution:

[HCl] = n/V = 0.048/0.1 = 0.48 M

According to the dissociation equation:

HCl + H2O --> Cl- + H3O+

the concentration of H3O+ ions is also 0.48 M, and:

pH = -log[H3O+] = -log0.48 = 0.32 approx.

Answer 2

my husband would know and he isn't here to ask....sorry!