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Neutralization reaction, pH of mixed solution of acid and base

ÀÛ¼ºÀÚ Uploader : ¹°¸®È­ÇÐ ÀÛ¼ºÀÏ Upload Date: 2018-09-20º¯°æÀÏ Update Date: 2018-09-20Á¶È¸¼ö View : 444

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Calculating pH of mixture of two solutions after neutralization reaction as below.

¡à acid solution
  - pH : ApH ( 0 ¡Â ApH < 7)
  - volume : VA (Liter)

¡à basic solution
  - pH : BpH ( 7 < BpH ¡Â 14 )
  - volume : VB (Liter)


¢º Input data


¡á Calculating the molality of H+ in acid solution

ApH = -log[H+] m, then,  [H+] = 10^(-ApH) = A (mol/L)

Molality of H+ in the solution : HVA = A * VA


¡á Calculating the molality of OH- in basic solution

pOH = 14 - BpH and pOH = -log[OH-], then, [OH-] = 10^(-(14-BpH)) = B (mol/L)

Molality of OH- in the solution : OHVB = B * VB


After the neutralization reaction below, there remains ions of either H+ or OH- whichever is the more.

H+(aq) + OH-(aq) ¡æ H2O(l)

¡á If H+ is more ( HVA > OHVB ),

volume of mixture of solutions is VA + VB. Thus, the molality of H+ ions is :




Then, pH of the mixture is :




¡á if OH- is more ( HVA < OHVB ),

Calculating the molality of OH- in the mixture,




This, pH of the mixture is :




If HVA = OHVB,  pH = 7
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