What is the pH?

Question

What is the pH of the solution formed when 35 mL of 0.22 mol/L HCl are mixed with 55 mL of 0.15 mol/L Mg(OH)₂.

Answer 1

First, write the neutralization reaction:

2HCl + Mg(OH)₂ ==> MgCl₂ + 2H₂O and pay attention to the coefficients (mole ratios)

We next want to find out if H⁺ or OH^- is present in excess. This will allow us to find the pH.

moles HCl present = 0.035 L x 0.22 mol/L = 0.0077 moles HCl

moles Mg(OH)₂ present = 0.055 L x 0.15 mol/L = 0.00825 moles Mg(OH)₂

Since it takes 2 mol HCl for each mol Mg(OH)2, clearly all the HCl will be used up, leaving some OH- behind.

Next we find how much OH is left:

0.0077 mol HCl x 1 mol Mg(OH)₂ / 2 mol HCl = 0.00385 mol Mg(OH)₂ used up to react with the HCl

moles Mg(OH)₂ remaining = 0.00825 - 0.00385 = 0.0044 moles Mg(OH)₂ left over.

Final volume is 35 ml + 55 ml = 90 ml = 0.090 L

Final [Mg(OH)₂] = 0.0044 mol/0.090 L = 0.0489 M

Since there are 2 OH- for each Mg(OH)₂, final [OH-] = 2 x 0.0489 = 0.0978 M OH-

pOH = - log [OH-] = -log 0.0978

pOH = 1.00

pH = 14 - 1

pH = 13

Answer 2

To start make sure all parts are in the same terms that is cm, mL, etc.

So 35 mL= (35ml/1000mL)(1 L)= 0.035 L and 55 mL= (55 mL/1000mL)(1 L)= 0.055 L respectively for the given concentrations.

Now remember pH is the measure of the concentration of H + ions.

To solve simply break each solution into it's respective components. Then find values pertaining to each component like you would to combine terms in foiling in Algebra or F_net in Physics do component wise. Then combine like terms to form parts of the new solution (that is mixture of the initial 2 solutions). Then use the formula for molality accordingly.

This is the English version breakdown of this problem. Do the Math according to your formulas given to you for this problem.