a solution of OH- ions is added to a beaker of pure water at 25 C, to a final concentration of 872 nM. Calculate the pH of the solution. I know that pH=-log[H+], but I'm still really confused as to how to convert the nM to M and how to solve this problem.

I assume 872 nM is 872 nanomolar which would make it 872-9M or 8.72E-7M. The problem states that OH is ADDED UNTIL the OH^- is 8.72E-7M. Now, since (H^+)(OH^-) = 1E-14

Plug in OH and solve for H. I get 1.15E-8 M and convert that to pH.

To solve this problem, you need to convert the concentration from nanomoles per liter (nM) to moles per liter (M).

To convert from nM to M, you need to divide the concentration by 1 billion (10^9), since there are 1 billion nanomoles in a mole.

Here's how to do the conversion:

1. Divide the concentration by 1 billion (10^9):
872 nM ÷ 10^9 = 8.72 × 10^-7 M

Now that you have the concentration in moles per liter (M), you can calculate the pOH using the formula pOH = -log[OH-].

Given that the concentration of OH- ions is 8.72 × 10^-7 M, we can proceed to calculate the pOH:

pOH = -log[OH-] = -log(8.72 × 10^-7) ≈ 6.06

Since pH and pOH are related by the equation pH + pOH = 14, we can calculate the pH:

pOH = 6.06
pH = 14 - pOH ≈ 14 - 6.06 ≈ 7.94

Therefore, the pH of the solution is approximately 7.94.