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2) a student standardized a naoh solution using h₂c₄h₄o₄. three h₂c₄h₄o…

Question

  1. a student standardized a naoh solution using h₂c₄h₄o₄. three h₂c₄h₄o₄samples were dissolved in water and titrated with naoh to a phenolphthalein end point. the titration reaction is shown in equation 9. equation 9: h₂c₄h₄o₄ (aq) + 2naoh (aq) → na₂c₄h₄o₄ (aq) + 2h₂o masses of h₂c₄h₄o₄ and naoh solutions required for titration are recorded in the table below. do the calculations for each titration and enter your answers in the table. a. calculate the number of moles of h₂c₄h₄o₄ titrated.

Explanation:

Step1: Calculate molar mass of \(H_2C_4H_4O_4\)

The molar mass \(M\) of \(H_2C_4H_4O_4\) (where \(H = 1\space g/mol\), \(C=12\space g/mol\), \(O = 16\space g/mol\)):

$$M=(2\times1)+(4\times12)+(4\times16)+(4\times1)=116\space g/mol$$

Step2: Calculate moles of \(H_2C_4H_4O_4\) for each determination

For Determination 1:

$$n_{H_2C_4H_4O_4}=\frac{m}{M}=\frac{0.2656\space g}{116\space g/mol}=0.00229\space mol$$

For Determination 2:

$$n_{H_2C_4H_4O_4}=\frac{m}{M}=\frac{0.2941\space g}{116\space g/mol}=0.002535\space mol$$

For Determination 3:

$$n_{H_2C_4H_4O_4}=\frac{m}{M}=\frac{0.2764\space g}{116\space g/mol}=0.002383\space mol$$

Answer:

Determination 1: \(0.00229\space mol\)
Determination 2: \(0.002535\space mol\)
Determination 3: \(0.002383\space mol\)