1) What is the pH-value of a buffer formed from 5.00 mL of 0.500 M acetic acid a
ID: 563862 • Letter: 1
Question
1) What is the pH-value of a buffer formed from 5.00 mL of 0.500 M acetic acid and 10.00 mL of 0.100 M NaOH solution? 2) A student needs 0.500 L of a 0.250 M sulfuric acid solution. How much of the concentrated sulfuric acid (98%) does he need? Show calculations. 1) What is the pH-value of a buffer formed from 5.00 mL of 0.500 M acetic acid and 10.00 mL of 0.100 M NaOH solution? 2) A student needs 0.500 L of a 0.250 M sulfuric acid solution. How much of the concentrated sulfuric acid (98%) does he need? Show calculations. 2) A student needs 0.500 L of a 0.250 M sulfuric acid solution. How much of the concentrated sulfuric acid (98%) does he need? Show calculations.Explanation / Answer
1)
CH3COOH= 5.00ml of 0.500M
number of moles of CH3COOH= 0.500M x 0.005L= 0.0025 moles
NaOH=10.00ml of 0.100M
number of moles of NaOH= 0.100M x 0.010L= 0.001 moles
CH3COOH + NaOH ------------------ CH3COONa + H2O
0.0025 0.001 0
-0.001 -0.001 +0.001
0.0015 0 0.001
after reaction , i,e after formation of buffer
number of moles of CH3COOH = 0.0015 moles
numberr of moles of CH3COONa = 0.001 moles
PKa of CH3COOH = 4.75
PH= PKa+ log[salt]/[acid]
PH= 4.75 + log(0.001/0.0015)
PH=4.573
PH= 4.57.
2) Volume = 0.500L = 500 ml
Molarity = 0.250M
molar mass of H2SO4 = 98 gram/mole
Molarity= maas/molar mass x1000/volume in ml
mass= Molarity x molar mass x volum in ml/1000
mass = 0.250 x 98 x 500/1000 =12.25 grams
Mass of H2SO4 = 12.15 grams.
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