The pH when 15 ml of 0.26 m HCl is titrated with 45 ml of 0.10 m NaOH is 13.35.
HCL + NaOH » NaCl + H2O
Given
Volume of NaOH = 45 ml = 0.045 l.
Volume of HCl = 15 ml = 0.015 l.
Concentration of NaOH = 0.26 m
Concentration of HCL = 0.10 m
Number of moles = E × V
Where,
C = Concentration
V = Volume
Number of moles of HCl = 0.10 × 0.015 = 0.0015 mol
Number of moles of NaOH = 0.045 × 0.26
= 0.0117 mol
Remaining moles of NaOH = 0.0117 – 0.0015 = 0.0102 mol
Molarity of NaOH = 0.0102 / 0.045 = 0.22 m
[OH] = 0.22M
pOH = -log[OH-]
pOH = -log(0.22)
pOH= 0.65
As we know that
pH +pOH= 14
pH = 14-pOH
pH = 14-0.6
pH= 13.35
Thus, we found that the pH is 13.35 if 15 ml of 0.26 m HCl is titrated with 45 ml of 0.10 m NaOH.
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