## Question based on Titration and pH computation

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Dhamnekar Winod
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### Question based on Titration and pH computation

Dissolve 2.0 g of the unknown monobasic acid sample in 100 ml of water. A 20 ml portion of this solution requires 15 ml of 0.12 M NaOH solution to reach the titrations equivalence point. If the molecular mass of the acid is 122 g/mol, determine the purity (%) of the acid.

How to answer this question? What is the answer to this question? I am working on this question. Any chemistry help, hint or even correct answer will be accepted.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
ChenBeier
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### Re: Question based on Titration and pH computation

Compare the theoretical consumption with the real one.
Calculate the moles you used and the moles of the NaOH react.
Dhamnekar Winod
Sr. Staff Member
Posts: 192
Joined: Sat Nov 21, 2020 10:14 am
Location: Mumbai[Bombay],Maharashtra State,India

### Re: Question based on Titration and pH computation

Solution:

The titration calculations for NaOH:

For 20 ml acid solution: 15 ml 0.12 mol NaOH required

So, the number of base equivalents = 0.12 × 0.015 = 1.8 × 10⁻³ equivalent

So, in 20 ml of acidic solution 1.80 x 10⁻³ equivalent of acids

Therefore molarity of acid = 9 × 10⁻³ equivalent,

In a 2 g sample:

Acid mass = 9 × 10⁻³ × 122 = 1.098 g

% Purity = 1.098/2 × 100 = 54.9%
This is weak acid strong base titration.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
ChenBeier
Distinguished Member
Posts: 873
Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany

Correct.