Question

2. Write the balanced chemical equation for the neutralization of hypobromous acid with potassium hydroxide. Is the pH at the equivalence point above, below, or exactly 7? Explain. 3. Some pH meters are designed for a three-point calibration at pH 4, 7, and 10. Ours are only calibrated with a two-point procedure at 4 and 7 or 7 and 10. Which range would you expect we are calibrating them at for this experiment? Why? 4. (a) An HCl solution is labeled 0.100 M. How is the pH of this solution expressed?

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Answer 1

2. The balanced reaction is HBrO + KOH KBro +H₂O In a neutralization reaction when an acid and a base react to form water and salt. and molves the combination of H + Cons and 1 OHO Cons to genegate Water. The neutralization of a strong acid and strong base has a pH equal to 7. And strong acid and weat base I will have a pH less than T. and conversely the strong base and weak adid the pH will be greater than to

So, In the given reaction KOH esa Strong base and HBrO ks a weak ach So the pH will be above 7. 3. The pH of the experiment should be 7 and 10 above 7 So 4 would expect to range pH metegs to be used. 4. a Ha H+ + că pH = -log[++] H7 = SHOLT = 0.1 pH – 109 10.1) b) pH of the o.IM Ha solution will be pH = - 109 0:1) 2lll c) When the sample is tested reads 0.87, where et and pH Meter should read !

It as showing the eanor of +0.13. The problem must be with the instrument. calibration should be done and correction should be done i,e acceptable errors. d) of the sample shows 2.94 instead of 1. where the value es completely very far from the original value a would like to follow the techniques below. 1. calibration of apparatus. 2. Control determination. 3. Blank determination 4. parallel determination. The performing the above techniques 9 will find the eggor and add enor to the test value 5. NaOH + Hal Nach + H2O b) No. of moles of HQ = 20x10x 0.10972 = 2.194 x 10 ml - volume intreg x molarith -3

c) My = M2 Va at equivalence point Imok of Hel and Imole of Naoh should react according to the reaction. so, Hal = 2.194 x 103 moles NaOH - 2.194 x 10 molesu,