Asked by Dexter
                Hi there - I'm going to post the question as written in my book. 
Three 20.0 mL samples of nitric acid are titrated with a 0.200 mol/L solution of sodium hydroxide. The results for the three trials are shown in the table below. What is the concentration of the hydrochloric acid?
Trial | 1 | 2 | 3 | average
_______________________________________
Final Burette| | | |
reading (ml) |28.20|27.90|28.00|
_______________________________________
Initial Buret|50.0 |50.0 |50.0 |
te reading(ml)
_______________________________________
Volume of
NaOH added | | | |
_______________________________________
end of question in book
my dilema:
Right off the bat I'm confused because I get a negative number for the volume added. ie. 28.20 ml - 50 ml = -21.80 ml which can't be right.
From there the rest of my answer is based on the negative number which I don't think is right. Please help me clarify how to work out the volume of the NaOH added if the final burette reading is 28.20 ml and initial burette reading is 50 ml. From there I think I can manage the rest but will probably request you verify
Thanks in advance!!!
 
            
            
        Three 20.0 mL samples of nitric acid are titrated with a 0.200 mol/L solution of sodium hydroxide. The results for the three trials are shown in the table below. What is the concentration of the hydrochloric acid?
Trial | 1 | 2 | 3 | average
_______________________________________
Final Burette| | | |
reading (ml) |28.20|27.90|28.00|
_______________________________________
Initial Buret|50.0 |50.0 |50.0 |
te reading(ml)
_______________________________________
Volume of
NaOH added | | | |
_______________________________________
end of question in book
my dilema:
Right off the bat I'm confused because I get a negative number for the volume added. ie. 28.20 ml - 50 ml = -21.80 ml which can't be right.
From there the rest of my answer is based on the negative number which I don't think is right. Please help me clarify how to work out the volume of the NaOH added if the final burette reading is 28.20 ml and initial burette reading is 50 ml. From there I think I can manage the rest but will probably request you verify
Thanks in advance!!!
Answers
                    Answered by
            DrBob222
            
    I'm confused also. Perhaps the buret is calibrated with 50.0 at the top and zero at the bottom so you have added 50-28.2 = 21.8 mL. However I've never seen a buret calibrated that way. Furthermore, it doesn't make any sense to calibrate it in reverse. It makes more sense for those 50.0 mL initial readings to be 0.50 or 0.05 or something like that.
    
                    Answered by
            Dexter
            
    That's what I thought....Thank you !!!
    
                    Answered by
            Vikas
            
    Experiment:-determine the normaliyy and normality and strength of the given solution of sodium hydroxide provided n/10 hcl  general calcutions normality and strength intial reading is 3 S.no is 0 and the final reading is 9,9.4 9.5 and volume of hcl used ? 
    
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