1) A 21.30 mL volume of 0.0975 M NaOH was used to titrate 25.0 mL of a weak mono

Question

1)

A 21.30 mL volume of 0.0975 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.

2.08 M

0.114 M

0.0831 M

0.00390 M

2.44 M

2)

For a weak acid (CH3COOH) that is titrated with a strong base (NaOH), what species (ions/molecules) are present in the solution at the stoichiometric point?

CH3COO-

H2O

Na+

NaCl

HCl

2.08 M

0.114 M

0.0831 M

0.00390 M

2.44 M

2)

For a weak acid (CH3COOH) that is titrated with a strong base (NaOH), what species (ions/molecules) are present in the solution at the stoichiometric point?

CH3COO-

H2O

Na+

NaCl

HCl

Explanation / Answer

1)

we have the Balanced chemical equation as:

NaOH + HA ---> NaA + H2O

Here:

M(NaOH)=0.0975 M

V(NaOH)=21.3 mL

V(HA)=25.0 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of HA

1*M(NaOH)*V(NaOH) =1*M(HA)*V(HA)

1*0.0975*21.3 = 1*M(HA)*25.0

M(HA) = 0.0831 M

Answer: 0.0831 M

2)

CH3COOH and NaOH react to form CH3COONa and H2O

CH3COONa is CH3COO- and Na+

Answer:

CH3COO-

Na+

H2O

Feel free to comment below if you have any doubts or if this answer do not work

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