Part I: Strong Acid/ Strong Base Measured p Calulated pH way Readun Strong Acid

Question

Part I: Strong Acid/ Strong Base Measured p Calulated pH way Readun Strong Acid Strong Base caluule tee all Mixed solution 3.IO Part II Phthalic acid buffer (KHP Calulated pH Ice Measured pH Phthalic acid solution Grams of Phthalic acid used 0,3 23 Moles of Phthalic acid 0.823 x mm ug Molarity of Phthalic acid solution produced Volume of Phthalic acid solution used to make buffer 0.010 L Molarity of NaOH solution 0.100 M 3 SF o.020 L Volume of NaOH solution used to make buffer Moles of NaOH used to make buffer 0,002 mates Acid Base Species molarity! Molarity of protonated weak acid hmolarity of conjugate base species in the buffer solution formed, (don't forget the change in molarity due to dilution and neutralization) Ypaet Calculated pH Measured pH pH of phthalic acid buffer: pH pKa log Acetic acid buffer Calculated pH lce. Measured pH Acetic acid solution Molarity of acetic acid solution Volume of acetic acid solution used to make buffer Moles of acetic acid present before mixing Molarity of NaOH solution 0.020L Volume of NaOH solution. Moles of NaOH added Acid Base Species molarity! Molarity of protonated weak acid /molarity of conjugate base species in the buffer solution formed (don't forget the change in molarity due to dilution and neutralization Measured pH Calculated pH pH of acetic acid buffer: pH pKa og

Explanation / Answer

Buffer

Phthalic acid

mass of phthalic acid = 0.823 g

moles of phthalic acid = 0.823 g/166.14 g/mol = 0.005 mol

Molarity of phthalic acid stock solution = 0.005 mol/0.05 L = 0.1 M

[*We have taken volume of solution 0.05 L for preparation of phthalic acid solution, If this value changes the molarity would change accordingly]

Volume of phthalic acid stock solution taken to prepare buffer = 0.04 L

moles of phthalic acid in this solution = 0.1 M x 0.04 L = 0.004 mol

Volume of 0.1 M NaOH taken to prepare buffer = 0.02 L

moles of NaOH = 0.1 M x 0.02 L = 0.002 mol

moles of NaPhthalate formed = 0.002 mol

molarity of NaPhthalate solution = 0.002 mol/0.06 L = 0.033 M

moles of phthalic acid remained = 0.002 mol

molarity of Phthalic acid solution = 0.002 mol/0.06 L = 0.033 M

pH = pKa + log([Naphthalate]/[phthalic acid])

     = 3.5 + log(0.033/0.033)

     = 3.5

Acetic acid

molarity of acetic acid solution = 0.1 M

Volume of acetic acid solution taken to prepare buffer = 0.04 L

moles of acetic acid before mixing = 0.1 M x 0.04 L = 0.004 mol

molarity of NaOH solution = 0.1 M

Volume of NaOH solution taken to prepare the buffer = 0.02 L

moles of NaOH before mixing = 0.1 M x 0.02 L = 0.002 mol

moles of sodium acetate formed = 0.002 mol

pH = pKa + log(sodium acetate formed/acetic acid remained)

      = 4.74 + log(0.002/0.002)

      = 4.74

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---