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a 5.172 g sample of a solid, weak, monoprotic acid is used to make a 100.0 ml solution. 35.00 ml of the resulting acid solution is then titrated with 0.09701 m naoh. the ph after the addition of 25.00 ml of the base is 5.32, and the endpoint is reached after the addition of 47.04 ml of the base.(a) how many moles of acid were present in the 35.00 ml sample

Respuesta :

bharathparasad577

In titration of an unknown weak monoprotic acid,  equivalence point is reached when the acid (analyte) is completely neutralized by  base (analyte).

One mole of  acid produce one mole of acetate; so,

0.025 L acid x 0.1 mol base = 0.0025 mol

1 L soln

0.0025 mol  x 1 mol = 0.0025 mol

1 mol acid

So, what volume is the acetate i will, Not 35 mL you added a solution

of acid

how much base did we add? 1 mol acid for the 1 mole of  base

0.0025 mol CH₃CO₂H x 1 mol NaOH = 0.0025 mol NaOH

1 mol CH₃COOH

where did the NaOH come from? a 0.1 M NaOH solution

0.0025 mol of the  NaOH x 1 L of solution = 0.025 L of the NaOH

0.1 mol NaOH

so, total volume is

total voumel = volume of acid + vol titrant

vol = 0.025 L + 0.025 L = 0.050 L

So, the concentration is

0.0025 mol of acid

= 0.050 M

0.050 L

To know more about pH of solution , please refer:

https://brainly.com/question/172153

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