Quentin Quickdigits has 28 25 mL of a sodium hydroxide solution of concentration
ID: 1036924 • Letter: Q
Question
Quentin Quickdigits has 28 25 mL of a sodium hydroxide solution of concentration 0 2528 M. He adds 14.63 mlL of a hydrochloric acid solution of concentration 0 1773 goal is to figure out the pH of the sodium hydroxide solution before and after addition of acid Calculate the initial pH (only two decimal places) before adding any HCI Answer Check Note this question does not ask for n, n, (before), n, (after). pOH. You vwill need to use those numbers to answer this question What is the final pH (two decimal places) of the mixture of the NaOH and HC1? Answer Check How many total ml of acid would be needed to reach the equivalence point? Answer Check Is the pH expected to be 7.00 at the equivalence point for all strong acid strong base titrations? Choose the best answer Select one o a. No it depends on whether you titrate with acid or base b.Yes, because the product is water Oc. Unable to determine from the information given O d No,each acid has a different K, o e. Yes, because it forms NaClExplanation / Answer
NaOH + HCl -----> NaCl + H2O
Initial pH before adding HCl is
molarity of NaOH = 0.2528
pOH = -log[0.2528] = 0.6
pH = 14-0.6 = 13.4
2) Final pH :
No of millimoles of NaOH = 0.2528 x 28.25 = 7.14
No. of millimoles of HCl = 14.63 x 0.1773 = 2.59
No. of millimoles of NaOH left after reaction = 7.14-2.59= 4.55
total vol = 28.25+14.63 = 42.88 ml
Molarity of Final NaOH solution = (4.55/42.88)= 0.106
pOH = -log[OH-] = -log [0.106] = 0.975
pH = 14-pOH = 14-0.975 = 13.025
3) Vol of acid needed for equivalence point V2
V1 = 25.28 ml M1 = 0.2528 M2 = 0.1773
M1V1 = M2V2
0.2528X 25.28 = 0.1773 X V2
V2 = 36.05 ml
4) pH is expected to be 7.00 at equivalence point for all stron acids and strong bases because water is formed by the reaction of proton and hydroxide ion.