chem - titration
2007-02-23 3:07 am
2.65g of anhydrous sodiuum carbonate was dissolved in water and made up to 250.0cm3 solution. 25.0cm3 of this solution required 26.3 cm3 of a hydrochloric acid solution for complete neutralization. Find the the molarity of the hydrochloric accid.
回答 (1)
2007-02-25 3:34 am
✔ 最佳答案
No.of moles of sodium Carbonate: 65/[23+23+12+16+16+16]=65/106No. of moles of sodium Carbonate in 25.0cm3 of this solution ={65/106} * {25/250}
=65/1060=13/212
By the equation: one mole of sodium Carbonate requires 2moles of hydrochloric acid to neutralise.
So: No of moles of hydrochloric acid present in the solution= 26/212
Let M be the molarity
26/212=26.3/1000 *(m)
M=4.66
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