What is the pH of a 6.40 × 10-3 M solution of sodium hydroxide, NaOH?

Question

anonymous · Answer

What is the pH of a 6.40 × 10-3 M solution of sodium hydroxide, NaOH?

anonymous · Answer

If 20.0 mL of a 0.500 M HCl solution is required to neutralize 35.0 mL KOH, what is the concentration of the KOH solution?

anonymous · Answer

For the first problem, you should take [OH-] = 6.40 x 10^(-3) since NaOH will dissociate completely.  

Then you can find pOH = -log[6.40 x 10^(-3)] = 2.1938

After you have pOH, plug in to:  pOH + pH = 14, and solve for pH.
2.1938 + pH = 14
pH = 11.8, a basic solution.  This makes sense because you have a fairly concentrated base in NaOH.
========================================
For the second problem, you need to use the formula $$M _{1}V _{1} = M _{2}V _{2}$$, 
where 1 is represented by HCl and 2 is represented by KOH:

(0.500 M)(20.0 ml) = x(35.0 ml)
x = 0.286 M, a lower concentration than the HCl, which makes sense because you have more of the KOH to begin with.

anonymous · Answer

Consider the following reaction:

4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)