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# 簡單的化學習題(需要詳細答案)

(1)

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)?

(2)

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).

(3)

(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.

Update:

Rating

(1)

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

(2)

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)

(3)

(a)

Na2S & H2O <--> NaHS & Na+ & OH-

NaHS & H2O <--> H2S & Na+ & OH-

(b)

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

= 0.1

Source(s): 腦內記憶,忘記的詞or公式就用搜尋
• 第二題可以再寫詳細一點嗎

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