A 50cm3 sample of solution which is 0.1M in HF is treated with 20cm3 of 0.1M NaOH. the concentration of HF remaining in the solution after addition of NaOH

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Answer 1

HF + NaOH ==> NaF + H2O  balanced equation

moles HF present = 0.05 L x 0.1 mol/L = 0.005 moles HF

moles NaOH added = 0.02 L x 0.1 mol/L = 0.002 moles NaOH

Since from the balanced equation you can see that they react in a 1 : 1 ratio, this means that 0.002 moles of HF have reacted with 0.002 moles NaOH, thus leaving 0.005 moles - 0.002 moles = 0.003 moles HF.

 

The CONCENTRATION of HF remaining is calculated from the moles (0.003) and the final volume which is 0.07 L. (0.05 L + 0.02 L = 0.07 L)

 

Final concentration of HF = 0.003 moles/0.07 L = 0.043 M = 0.04 M (to 1 significant figure)