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1. write the balanced chemical equation \\( \\mathrm { hc } _ { 2 } \\m…

Question

  1. write the balanced chemical equation

\\( \mathrm { hc } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } ( a q ) + \mathrm { naoh } ( a q ) \
ightarrow \mathrm { h } _ { 2 } \mathrm { o } ( l ) + \mathrm { nac } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } ( a q ) \\)

  1. calculate moles of sodium hydroxide

\\( \mathrm { mol } \mathrm { naoh } = 22.76 \mathrm { ml } \mathrm { naoh } \times \frac { 1 \mathrm { l } } { 1000 \mathrm { ml } } \times \frac { 0.1026 \mathrm { mol } \mathrm { naoh } } { 1 \mathrm { l } \mathrm { naoh } } = 2.335 \times 10 ^ { - 3 } \mathrm { mol } \mathrm { naoh } \\)

  1. calculate moles of acetic acid

\\( \mathrm { mol } \mathrm { hc } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } = 2.335 \times 10 ^ { - 3 } \mathrm { mol } \mathrm { naoh } \times \frac { 1 \mathrm { mol } \mathrm { hc } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } } { 1 \mathrm { mol } \mathrm { naoh } } = 2.335 \times 10 ^ { - 3 } \mathrm { mol } \mathrm { hc } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } \\)

  1. molarity of of acetic acid

\\( \left \mathrm { hc } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } \
ight = \frac { 2.335 \times 10 ^ { - 3 } \mathrm { mol } \mathrm { hc } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } } { 10.00 \mathrm { ml } \mathrm { hc } _ { 2 } \mathrm { h } _ { 3 } \mathrm { o } _ { 2 } } \times \frac { 1000 \mathrm { ml } } { 1 \mathrm { l } } = 0.2335 \mathrm { m } \\)

Explanation:

Step1: Write balanced chemical equation

$$HC_2H_3O_2(aq)+NaOH(aq)\to H_2O(l)+NaC_2H_3O_2(aq)$$

Step2: Calculate moles of NaOH

$$n_{NaOH}=22.76\ mL\times\frac{1\ L}{1000\ mL}\times\frac{0.1026\ mol}{1\ L}=2.335\times10^{-3}\ mol$$

Step3: Calculate moles of acetic acid

From the balanced equation, mole ratio \(HC_2H_3O_2:NaOH = 1:1\)
$$n_{HC_2H_3O_2}=n_{NaOH}=2.335\times 10^{-3}\ mol$$

Step4: Calculate molarity of acetic acid

$$M=\frac{n}{V}=\frac{2.335\times 10^{-3}\ mol}{10.00\ mL\times\frac{1\ L}{1000\ mL}} = 0.2335\ M$$

Answer:

The molarity of acetic acid is \(0.2335\ M\)