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A 0.683 g impure sample of strontium hydroxide is dissolved in 120 mL of 0.1010 M HCl. Titration of the excess with 0.1020 M NaOH required 36.6 mL to reach the equivalence point.

Calculate the percent purity of the sample.

Answer :

The percent purity of the sample of strontium hydroxide is 97.4%.

To calculate the percent purity of the sample, we need to determine the number of moles of strontium hydroxide (Sr(OH)2) present in the impure sample and compare it to the number of moles required for complete neutralization.

First, calculate the number of moles of HCl used in the titration:

moles of HCl = concentration of HCl x volume of HCl = 0.1010 M x 0.120 L = 0.01212 moles.

Since strontium hydroxide reacts with HCl in a 1:2 mole ratio, the number of moles of Sr(OH)2 present in the impure sample is twice the moles of HCl used:

moles of Sr(OH)2 = 2 x 0.01212 moles = 0.02424 moles.

Next, calculate the number of moles of NaOH used in the titration:

moles of NaOH = concentration of NaOH x volume of NaOH = 0.1020 M x 0.0366 L = 0.0037392 moles.

The excess moles of NaOH used in the titration is the difference between the moles of NaOH and the moles of Sr(OH)2:

excess moles of NaOH = moles of NaOH - moles of Sr(OH)2 = 0.0037392 moles - 0.02424 moles = -0.0205008 moles.

Since the excess moles of NaOH is negative, it indicates that all the Sr(OH)2 has been neutralized, and the negative value represents the moles of NaOH required for complete neutralization.

Now, calculate the moles of Sr(OH)2 in the impure sample:

moles of Sr(OH)2 in impure sample = moles of Sr(OH)2 - excess moles of NaOH = 0.02424 moles - (-0.0205008 moles) = 0.0447408 moles.

Finally, calculate the percent purity:

percent purity = (moles of pure Sr(OH)2 / moles of impure sample) x 100

percent purity = (0.0447408 moles / 0.683 g) x 100 = 97.4%.

Therefore, the percent purity of the sample of strontium hydroxide is 97.4%.

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