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Which one of the following salts forms aqueous solutions with pH = 7? Na_2S NaBr

ID: 502627 • Letter: W

Question

Which one of the following salts forms aqueous solutions with pH = 7? Na_2S NaBr NaCIO_2 NaNO_2 Na_2CO_3 The pH of an aqueous sodium fluoride (NaF) solution is ___ because ___ 7; sodium fluoride is a simple salt. above 7; fluoride is a weak base, below 7; fluoride reacts with water to make hydrofluoric acid, about 7; fluoride is a weak base, but produces hydrofluoric acid, and these two neutralize one another. 0; sodium fluoride is a salt not an acid or a base. What is the pH of a solution in which [HA] = [A^-]? pH = 1 pH = K_a pH = pK_a pH = pOH pH = 7.0 Which of the following can be mixed together in water to produce a buffer solution? HCIO_4 and NaCIO_4 HNO_3 and NaNO_3 H_2SO_4 and NaHSO_4 H_3PO_4 and NaH_2PO_4 HCl and NaCl Acid-base indicators change color ___ exactly when pH = pK_a of the indicator. generally over a range of 1 or 2 pH units. at pH = 7. always between a pH of 6 and 8. at the midpoint of a titration. In a titration of monoprotic acids and bases, there is a large change in pH ___ at the point where pH = pK_a, of the acid. when the volume of acid is exactly equal to the volume of when the concentration of acid is exactly equal to the concentration of base. when the number of acid is exactly equal to the number of moles of base, at the point at which pH = p K_b of the base.

Explanation / Answer

51) ans = b

For salts formed from strong acid and strong base , pH = 7

pH of NaBr = 7

This is bacause , NaBr formes from string acid HBr and strong base NaOH.

53) ans = c

From Henderson-Hasselbalch equation,

pH = pKa + log [A-]/[HA]

For [HA] = [A-],

pH = pKa + log [A-]/[A-]

= pKa + log 1

= pKa + 0

= pKa

Therefore,

pH = pKa

54) ans = d

weak acid + its conjugate base = buffer

weak base + its conjugate acid = buffer

HClO4,HNO3,H2SO4 and HCl do not form buffer solitions because theyare strong acids.

H3PO4 is weak acis. So, it form buffer solution.

Therefore.

ans = d = H3PO4 and NaH2PO4

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