How many milliliter of 0.0195 M HCl are required to titrate
(A) 25mL 0.0365M KOH(aq) (B)10.00mL 0.0116M Ca(OH)2(aq) (C) 2mL 0.0225M NH3(aq)?
In a titration experiment , a student finds that 25.00 mL of NaOH solution are needed to completely neutralize 0.5105g of KHP .What is the concentration (in molarity) of the NaOH solution ?
(Mw of KHP :204.2g/mole).
(a)Wrire a balanced acid-base reaction that produces Na2S
(b)If Na2S is added to an aqueous solution of H2S (Ka=9.1*10-8),will the pH of the solution rise or fall ?Explain.
- 1 decade agoFavorite Answer
KOH mole數 = 25*0.0365 mmole
because it needs the equal mole of HCl
25*0.0365/0.0195 = 46.79 (mL)
(A)-----Ans: 46.79 ml
Ca(OH)2 mole數 = 10*0.0116 mmole
because it needs the 2 times of HCl for neutralize
(1 mole Ca(OH)2 produce 2 mole OH-)
10*0.0116 * 2 /0.0195 = 11.90 (mL)
(B)-----Ans: 11.90 ml
NH3 mole數 = 2*0.0225 mmole
because it needs the equal mole of HCl to neutralize
(NH3 with H2O --> NH4+ with OH- )
2*0.0225 /0.0195 = 2.31 (mL)
(C)-----Ans: 2.31 ml
KHP為單質子酸, just needs the same mole of NaOH to neutralize
mole of KHP = 0.5105 / 204.2 (mole)
mole of NaOH = 25/1000* concentration
-----Ans: [NaOH] = 0.1 (M)
Na2S & H2O <--> NaHS & Na+ & OH-
NaHS & H2O <--> H2S & Na+ & OH-
Of cource rise
becourse H2S is weak acid Na2S is strong base
strong base added into weak acid will that the pH volume rise
and in this case, it will become a buffer.
2009-03-30 18:37:25 補充：
KHP 每莫爾重204.2g (Mw 204.2g/mole)
0.5105g KHP含 KHP 0.5105 / 204.2 mole
NaOH使用量為 25mL = 0.025L
即 25ml NaOH 溶液中 有 NaOH <<0.5105 / 204.2>> mole(剛好中和KHP)
molarity為 溶質mole數 / 溶液容量 (Liter)
(0.5105 / 204.2) / 0.025
= 0.1Source(s): 腦內記憶,忘記的詞or公式就用搜尋
- 1 decade ago