Question
Asked Nov 3, 2019

A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoichiometric point. Determine the molar concentration of the weak acid solution. 

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Step 1

When acid reacts with a base, it leads to the formation of salt and water. The chemical reaction for the same is shown as follows:

НА + NaOH — NaA + H,O
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НА + NaOH — NaA + H,O

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Step 2

As per the data given in the question,

Volume of NaOH = 23.74 mL = 0.02374 L (As 1 mL = 10-3 L)

Concentration of NaOH = 0.0981 M

Therefore, the number of moles of NaOH will be calculated as follows:

moles of NaOH= volume x concentration of NaOH
0.02374 Lx 0.0981M
0.002328894 mol
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moles of NaOH= volume x concentration of NaOH 0.02374 Lx 0.0981M 0.002328894 mol

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Step 3

According to the chemical reaction mentioned above,

moles of HA = moles of NaOH = 0.002328894 mol

It is given that, volume of HA = 25.0 ...

moles of HA
concentration of HA =
volume of HA
0.002328894mol
0.025 L
0.09315 M
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moles of HA concentration of HA = volume of HA 0.002328894mol 0.025 L 0.09315 M

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