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 Post subject: Task! Chemistry.
PostPosted: Mon, 26 Dec 2011 19:37:16 UTC 
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How should I solve this one??!!


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 Post subject: Re: Task! Chemistry.
PostPosted: Tue, 27 Dec 2011 01:15:44 UTC 
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kreshnik wrote:


You need to know the pK_a value.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Task! Chemistry.
PostPosted: Tue, 27 Dec 2011 14:37:26 UTC 
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sorry... my fault :(

0.1 mol \:\: CH3COOH

[H^+]=?\:\:\:\:\:pH=?

pKa=4.75

Ka=1.8*10^{-5}


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 Post subject: Re: Task! Chemistry.
PostPosted: Wed, 28 Dec 2011 01:12:59 UTC 
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kreshnik wrote:
sorry... my fault :(

0.1 mol \:\: CH3COOH

[H^+]=?\:\:\:\:\:pH=?

pKa=4.75

Ka=1.8*10^{-5}


Remember we have the equilibrium \mathrm{{CH_3COOH}_{(aq)}}+\mathrm{{H_2O}_{(l)}} \rightleftharpoons\mathrm{{H_3O}^+_{(aq)}}+\mathrm{{CH_3COO}^-_{(aq)}}, and the acid dissociation constant K_a=\dfrac{[\mathrm{{H_3O}^+_{(aq)}}][\mathrm{{CH_3COO}^-_{(aq)}}]}{[\mathrm{{CH_3COOH}_{(aq)}}]}. Now solve a quadratic equation and hence \mathrm{pH}=-\log_{10}[\mathrm{{H_3O}^+_{(aq)}}].

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Task! Chemistry.
PostPosted: Wed, 28 Dec 2011 11:05:18 UTC 
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Joined: Sun, 25 Dec 2011 00:28:25 UTC
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Thanks a lot :D


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