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# 100 ml of 0.2 molar acetic acid with 50 ml of 0.1 M and magnesium Hydroxide calculate the pH of the final solution

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Solution

## Start by determining the concentrations of the two species in the buffer. For acetic acid, you have C=n/V⇒nacetic =C⋅V =0.2M⋅100⋅10^−3L =20⋅10^−3moles This means that the concentration of acetic acid in the buffer will be Cacetic=nacetic/Vbuffer =20⋅10^−3moles/(100+100)⋅10^- 3L =0.1 M For sodium acetate, the number of moles is nacetate=C⋅V =0.15xMx100⋅10^−3L =15⋅10−3moles As a result ,the concentration of MgOH in the buffer will be CMgOH=nMgOH/Vbuffer =10⋅10^−3moles/(100+50)x10^-3 = 0.6667 Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution pHsolution=pKa+log([MgOH][CH3COOH]) pHsolution=4.75+log(0.0667M ÷0.1M) =4.75−0.175=4.575

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