5. You add 11.00 mL of 0.800 M NaOH to 50.00 mL of pure water, and to this mixture you then add 3.00 mL of 0.200 M HCl. What will be the pH of the resulting solution

pH = 13.11

Explanation:

NaOH reacts with HCl to produce H2O and NaCl. pH depends of this reaction. The reactant in excess determine the pH of the solution:

NaOH + HCl → H2O + NaCl

1 mole of NaOH reacts with 1 mole of HCl

To determine limiting reactant we need to find moles of each reactant:

Moles NaOH:

0.011L * (0.800 mol / L) = 0.0088 moles NaOH

Moles HCl:

0.003L* (0.200mol / L) = 0.0006 moles HCl

That means limitng reactant is HCl and moles in excess of NaOH are:

0.0088moles NaOH - 0.0006 moles = 0.0082 moles NaOH =

0.0082 moles OH⁻

In 11.00mL + 50.00mL + 3.00mL = 64.00mL = 0.06400L

[OH-] = 0.0082 moles OH- / 0.064L = 0.1281M OH-

pOH = -log [OH-] = 0.89

pH = 14 - pOH

the answer is alpha particles hope i you

explanation:

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