Don\'t understand why my first answer is wrong can someone help me? eneral Chemi

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Don't understand why my first answer is wrong can someone help me? eneral Chemistry 4th Edition University Science Books presented by Sapling Leaning Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solutic that is 0.100 M in NH3(aq) and 0.100 M in NH4CI(aq). A list of ionization constants can be found here Number ApH-9.25 Calculate the change in pH when 3.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution Number ApH 0.026 There is additional feedback available! View this feedback by clicking on the bottom divider bar on the divider bar again to hide the additional feedback Close Previous 2? Try Again Next

Explanation / Answer

pH of salt solution is pH = pKa + log[salt]/[acid]

or pOH = pKb + log[salt]/[base]

pOH = 4.75 + log(100 * 0.1/100)/((100*0.1)/100))

pOH = 4.75

pH = 14 - pH = 14 - 4.75 = 9.25

by adding 3 mL of 0.1 M NaoH then,

pOH = pKb + log[salt - NaOH]/[Base + NaOH]

pOH = 4.75 + log(((100*0.1 - 3*0.1)/(100+3))/((100*0.1 + 3*0.1)/(100+3)))

pOH = 4.723

pH = 14 - 4.723

pH = 9.277

Change in pH = 9.277 - 9.25 = 0.027

by adding 3 mL of 0.1 M HCl then,

pOH = pKb + log[salt + HCl]/[Base - HCl]

pOH = 4.75 + log(((100*0.1 + 3*0.1)/(100+3))/((100*0.1 - 3*0.1)/(100+3)))

pOH = 4.776

pH = 14 - 4.776

pH = 9.224

Change in pH = 9.25 - 9.224 = 0.026

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