Originally posted by caroline86
5.267g anhydrous Na2CO3 was disolved in water in a volumetric flask and the volume made up 250mL.
n(Na<sub>2</sub>CO<sub>3</sub>) = m / M = 5.267 / (2 * 22.99 + 12.01 + 3 * 16.00) = 5.267 / 105.99 = 0.049693... mol
So, [Na<sub>2</sub>CO<sub>3</sub>] = n / V = 0.049693... / (250.0 * 10<sup>-3</sup>) = 0.19877... molL<sup>-1</sup>
in 10mL of this solution was pipetted into a conical flask and titrated with hydrochloric acid. 21.3mL was needed to reach the equivalence point. calculate the molarity of the hydrochloric acid solution.
Na<sub>2</sub>CO<sub>3</sub> + 2HCl ---> 2NaCl + CO<sub>2</sub> + H<sub>2</sub>O
n(Na<sub>2</sub>CO<sub>3</sub>) = CV = 0.19877... * 10.00 * 10<sup>-3</sup> = 1.9877... * 10<sup>-3</sup> mol
Now, from equation, n(HCl) / n(Na<sub>2</sub>CO<sub>3</sub>) = 2 / 1
So, n(HCl) = (2 / 1) * 1.9877... * 10<sup>-3</sup> = 3.9754... * 10<sup>-3</sup> mol
So, [HCl] = n / V = 3.9754... * 10<sup>-3</sup> / (21.30 * 10<sup>-3</sup>) = 0.18664... = 0.1866 molL<sup>-1</sup> (4 sig fig)
this solution was then used to determine the concentration of an unknown barium hydroxide solution. 25mL of the barium hydroxide solution required 27.1mL hydrochloric acid solution for exact neutralisation. calculate the molarity of the barium hydroxide solution. in addition, calculate its concentration in grams per litre.
what the??? how do you do this?
This is the same basic calculation, so you give it a go. The answer is [Ba(OH)<sub>2</sub>] = 0.1012 molL<sup>-1</sup>, and you can find the concentration in grams per litre using the formula:
Conc in gL<sup>-1</sup> = Conc in molL<sup>-1</sup> * Molar mass in gmol<sup>-1</sup>
