Question

A 5.00mL aliquot of H2SO4 is neutralized with 23.55mL of a 0.1000M NaOH solution.

a. calculate the pH of the acid.

b. calculate the % (w/v) of the acid.

Answer #1

We know from the law of acidimetry and alkalimetry,

V_{1}S_{1} = V_{2}S_{2} where,
V_{1}= volume of H2SO4 solution = 5 mL

S_{1} = strength of H2SO4 solution

V_{2}= volume of NaOH solution = 23.55 mL

S_{2}= strength of NaOH solution =0.1M

Therefore, strength of H2SO4 solution ( S_{1}
) = V_{2}S_{2} / V_{1} = 23.55 x
0.1 / 5 (M) = 0.471 (M)

(a) Then pH of H2SO4 solution = - log [H+] = - log [ 2 x 0.471 ] = - log 0.942 = 0.0259 (ANSWER)

(b) 1mole H2SO4 = 98.079 g

So, 0.471 mole H2SO4 = 98.079 x 0.471 = 46.195 g

Then 1000 mL of H2SO4 contain 46.195 g of acid

Therefore, 100 mL of H2SO4 contain { (46.195 x 100 ) /1000 } g = 4.6195 g of acid

So, % (w/v) of the acid is 4.6195 g /100 mL (ANSWER)

A
sample of acetic acid (weak acid) was neutralized with .05M NaOH
solution by titration. 34 mL of NaOH had been used. Show your work.
CH3COOH + NaOH-----CH3COONa + H2O
a) Calculate how many moles of NaOH were used?
b) How many moles of aspirin were in a sample?
c) Calculate how many grams of acetic acid were in the
sample
d) When acetic acid is titrated with NaOH solution what is the
pH at the equivalence point? Circle the...

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An unknown mass of NaOH is dissolved in 500g of H20. This
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The mass of acetic acid solution is 30.01 g
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The concentration of standardization NaOH titration is 0.09755
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** In my notes it says that the...

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c) after half the acid is neutralized
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