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[LSDP]

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I need help with this question. Thank you in advance.

#1. Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H2C2O4 ) in a solid given that a 0.7984 g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

A) 48.89
B) 97.78
C) 28.59
D) 1.086
E) 22.83

The difficult level is 5/5 😱. Please give me the explanation to your answer.

 

[lutopie0730]

I need help with this question. Thank you in advance.

#1. Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H2C2O4 ) in a solid given that a 0.7984 g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

A) 48.89
B) 97.78
C) 28.59
D) 1.086
E) 22.83

The difficult level is 5/5 😱. Please give me the explanation to your answer.

is the answer A?
if it is, then here it goes

H2C2O4 + 2NaOH -> 2H2O + Na2C2O4

(37.98ml)(0.2283M)= 8.671 mmol NaOH

(8.671 mmol)(1/2)= 4.336 mmol H2C2O4

(4.336 mmol H2C2O4)(90mg/mmol)=390.2 mg = 0.3902 g

(0.3902 / 0.7984)*100% = 48.87%

 

[LSDP]

You are correct. Thank you for the quick reply. The answer is A. Why multiple 1/2? My intuition tell me to multiple by 2 because of 2 protons.

 

[lutopie0730]

It is clear if you write down the conversion factors based on the balanced equation above.

(8.617 NaOH)(1 H2C2O4/2NaOH)=4.336 H2C2O4 -> NaOH is canceled out 🙂

 

[LSDP]

I get it now. I was doing it too fast that I forgot to visualize the equation and plus I only had 3hrs of sleep. Thanks again.