20cm3 of saturated aqueous solution of hydrogen carbonate (iv) at 25^oc requires 10cm3 of 2.0moldm-3 hydrochloric acid for complete reaction.what is the solubility of sodium hydrogen carbonate(iv) in moldm-3 at 25^oc
To find the solubility of sodium hydrogen carbonate (NaHCO3) in moldm-3 at 25°C, we need to determine the concentration of NaHCO3 in the saturated aqueous solution.
From the information given, we know that 20cm3 of the saturated solution of NaHCO3 reacted with 10cm3 of 2.0moldm-3 HCl (hydrochloric acid) for complete reaction.
Firstly, let's calculate the amount of hydrochloric acid (HCl) used in moles:
10 cm3 is equal to 10/1000 = 0.01 dm3
Moles of HCl = concentration (moldm-3) × volume (dm3)
Moles of HCl = 2.0 × 0.01 = 0.02 mol
Now, since the reaction between NaHCO3 and HCl is in a 1:1 ratio (one mole of NaHCO3 reacts with one mole of HCl), the number of moles of NaHCO3 is also 0.02 mol.
We are given the volume of the saturated solution as 20 cm3, which is equal to 20/1000 = 0.02 dm3.
Next, we can calculate the solubility of NaHCO3:
Solubility (moldm-3) = moles of NaHCO3 / volume of solution (dm3)
Solubility = 0.02 mol / 0.02 dm3
Solubility = 1.0 moldm-3
Therefore, the solubility of sodium hydrogen carbonate (NaHCO3) in moldm-3 at 25°C is 1.0 moldm-3.