Calculating pH of solutions?

PH of mixture of solutions, Is this right?

• What is the pH of a solution prepared by mixing 60.0 mL: of LiOH solution with pH = 12.28 and 40.0 mL of LiOH solution with pH = 13.19. okay need to get the moles of both of the soltutions, so first need to get the H+ concentrations of them. [H+] (1) = 10^-12.28= 5.2x10^-13M [H+] (2) = 10^-13.19= 6.5x10^-14M now that we have H+, we can convert them to moles by multiplying their Molarity with their Volume. nH (1) = MV = 5.2x10^-13 x .060L = 3.1x10^-14 nH (2) = MV = 6.5x10^-14x .040L = 2.6x10^-15 Now that we have both the moles we add them together and devide by their totatl volume in liters. ((3.1x10^-14) + (2.6x10^-15))/.100L = 3.4x10^-13 That gives the molarity of the mixed solutions, now to get the ph. -log[3.4x10^-13] = 12.47 pH This seems right as its inbetween the 2 ph's and it is basic, do you guys get the same answer, and did I mess up on any significant figs, I have trouble with those in long problems.

• Answer:

  Looks Good!