I have just stumbled upon a very weird question that I was able to do most of but got stuck at one step and the solution makes absolutely no sense. Please help me out. The question is as follows:
A standard solution was prepared by dissolving 2.628g of analytical grade sodium hydrogen carbonate in sufficient distilled water to give a final volume of 500.0 mL. The standard solution was then used to determine the concentration of hydrochloric acid. Four 25.00 mL samples of the acid were titrated with the standard sodium hydrogen carbonate solution. The titration results are shown below.
Titration 1 - Endpoint volume = 23.40 mL
Titration 2 - Endpoint volume = 23.55 mL
Titration 3 - Endpoint volume = 24.75 mL
Titration 4 - Endpoint volume = 23.35 mL
Determine the concentration of the hydrochloric acid solution.
Heya!!
Let's write out the equation
First, we can find the concentration of the original 500mL NaHCO3 solution made
 = \frac{2.628}{22.99+1.008+12.01+3\times 16} =0.03128...\text{ mol} \\ c(NaHCO_3 )=\frac{0.03128...}{0.5}=0.06256 \text{ mol/L} )
We can then find the volume of NaHCO3 solution used in the titration by averaging the endpoint volumes, and use this to find the moles of NaHCO3 that reacted in the titration
 = \frac{0.0234+0.02355+0.02475+0.02335}{4} = 0.0237625\text{ L} \\ n(NaHCO_3 )=0.06256\times 0.0237625 =0.001486582 \text{ mol})
Using the mole ratio, we can find the moles of HCl, and given that v(HCl)=0.025, we can find the concentration of HCl!!
=n(NaHCO_3 ) = 0.001486582 \text{ mol} \\ c(HCl) = \frac{0.001486582}{0.025}=0.05946 \text{ mol/L})
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