in Enthalpy for the reaction (interpret each values and uaing les Law Caleulate

Question

in Enthalpy for the reaction (interpret each values and uaing les Law Caleulate ehange HS-233 kj ake a buffer given the following dium acetate, a stock bottle of 5.OM acetic acid, water and lab equipment is available; ml; pH-4.60 ka 1.8 x 10The total moles of buffer: 0.300 moles 5.0 55.s How would you make a buffer given the following: Buwortle of solid sodium acetate, a stock bottle of 6.0Macetie acid, water and lab equipment is available :"500mL; pH-460 ka = 1.8 x 10s; Thetotal moles of buffer: 0.200moles Acid Bdse

Explanation / Answer

Answer:

We know the pH of the buffer is calculatd by using Hendersen equation.

pH = pKa + log [conjugate base]/acid]

Given

the pKa of acetic acid = 4.75

molarity = 6.0M

For the buffer

pH = 4.60

volume = 500mL and

total moles = 0.2 00

Let us take moles of conjugate base , acetate as x moles

then moles of acetic acid in solution =0.2-x

thus pH = 4.6 = 4.75 + log x/0.2-x

Thus x = 0.0829 mol

Thus moles of conjugate base in 500mL buffer = 0.0829 mol

Thus mass of sodium acetate needed to make this many moles = moles x molar mass

= 0.0829 mol x 82g/mol

= 6.7978g

NOw the moles of acetic acid in 500mL buffer = 0.200 - 00829 = 0.1171 moles

thus molarity of acetic acid in buffer solution = 0.1171/0.5L = 0.234 M

We need to dilute the stock solution of 6.0M acetic acid

for dilution

V1M1 = V2M2

500mL x 0.234M = 6.0M x V

Thus Volume of stock solution needed = 19.5 mL

To prepare the reuired uffer

take 19.5mL of 6.0M acetic acid solution in to a 500mL volumetric flask, transfer 6.7978 g of sodium acetate, dissolve in water and make up to mark. you have a buffer of pH = 4.6 with total moles of buffer = 0.200

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