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natima [27]
4 years ago
14

It takes 26.8 mL of a 0.0700 M NaOH standard solution to neutralize a 250 mL sample of lactic acid (C3H6O3). What mass of lactic

acid was dissolved in the sample?
Chemistry
1 answer:
makkiz [27]4 years ago
6 0

The complete balanced chemical reaction for this would be:

<span>NaOH  +  C3H6O3  -->  NaC3H5O3  +  H2O</span>

 

So we see that exactly 1 mole of NaOH reacts with 1 mole of lactic acid.

Calculate moles of NaOH.

moles NaOH = 0.07 moles/L * 0.0268 L

moles NaOH = 1.876 x 10^-3 mol

 

So,

moles C3H6O3 = 1.876 x 10^-3 mol

The molar mass of lactic acid is 90.08 g/mol so the mass is:

mass C3H6O3 = (1.876 x 10^-3 mol) * 90.08 g/mol

<span>mass C3H6O3 = 0.169 g</span>

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                                KH₂PO₄ + NaOH → NaKHPO₄ + H₂O
initial                         0.343       1.096W       0                 0
after reaction (0.343 - 1.096W)     0         1.096W        1.096W
KH₂PO₄ and NaKHPO₄ is considered as buffer and its (H₂PO₄⁻) pKa value is 7.21
phenolphthalein indicator turns pink means it is at the equivalence point
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by using buffer equation
pH = pKa + log [salt] / [acid]
when pH = pKa so [salt] = [acid]
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2.192 W = 0.343
W = 0.156 g
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