Calculate the ph of the resulting solution if 34.0 ml of 0.340 m hcl (aq is added to

According to the reaction equation:

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

a) Part 1):

first, we need to get moles HCl = molarity * volume

= 0.34m * 0.034

= 0.01156 mol

then, we need moles of NaOH = molarity * volume

= 0.34 * 0.039

= 0.01326 mol

when NaOH excess:

∴ NaOH remaining = 0.01326 - 0.01156

= 0.0017 mol

when the total volume = 0.039 + 0.034

= 0.073 L

∴[OH] = moles / total volume

= 0.0017mol / 0.073 L

= 0.0233 M

∴ POH = - ㏒[OH-]

= - ㏒0.0233

= 1.63

∴ PH = 14 - 1.63

= 12.37

b) part 2:

as we got moles HCl = molarity * volume

= 0.34 * 0.034

= 0.01156 mol

then moles NaOH = molarity * volume

= 0.39 * 0.044

= 0.01716 mol

NaOH remaining = 0.01716 - 0.01156

= 0.0056 mol

when the total volume = 0.034 + 0.044

= 0.078 L

∴[OH-] = 0.0056 / 0.078

= 0.0718 M

∴ POH = - ㏒[OH-]

= - ㏒0.0718

= 1.14

∴PH = 14 - POH

= 14 - POH

= 14 - 1.14

= 12.86