Consider the titration of a 23 4 mL sample of 0 125 MRbOH with0 110 M ...

A RbOH Rb OH 0125M 0125M OH RbOH OH 0125M POH log0125 09030 PH 14POH 1409030 13097 HCl ....

Consider the titration of a 23.4 âˆ’mL sample of 0.125 MRbOH with 0.110 M HCl. Determine...

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Consider the titration of a 23.4 âˆ’mL sample of 0.125 MRbOH with0.110 M HCl. Determine each of the following.
Part A) the initial pH
Part B) the volume of added acid required to reach theequivalence point
Part C) the pH at 5.7 mL of added acid
Part D) the pH at the equivalence point
Part E) the pH after adding 6.0 mL of acid beyond theequivalence point

Answer & Explanation Solved by verified expert

3.6 Ratings (674 Votes)

A. RbOH -------> Rb⁺ + OH^-

^{Â Â Â Â Â Â Â} 0.125MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.125M

Â Â Â [OH-]Â Â Â = [RbOH]

Â Â [OH-]Â Â = 0.125M

Â Â POHÂ Â = -log0.125

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.9030

PHÂ Â Â = 14-POH

Â Â Â Â Â Â Â Â Â Â = 14-0.9030Â Â = 13.097

Â Â HCl + RbOH --------> RbCl + H2O

no of moles of HCl = no of moles of RbOH

Â Â M1V1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = M2V2

0.11*V1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.125*23.4

Â Â Â Â Â Â Â Â V1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.125*23.4/0.11Â Â Â = 26.6ml

Â Â MÂ Â = MBVB -MAVA/VA+VB

Â Â Â Â Â Â Â Â Â Â = 0.125*23.4-0.11*5/23.4+5

Â Â Â Â Â Â Â Â Â Â = 2.375/28.4Â Â = 0.084M

[OH-] = M = 0.084M

POHÂ Â = -log[OH-]

Â Â Â Â Â Â Â Â Â Â Â = -log0.084Â Â = 1.0757

PHÂ Â Â = 14-POH

Â Â Â Â Â Â Â Â = 14-1.0757Â Â = 12.9243

D. MÂ Â = MAVA -MBVB/VA+VB
Â Â Â Â Â Â Â Â Â Â =0.11*26.6- 0.125*23.4/23.4+26.6

Â Â Â Â Â Â Â Â Â Â = 0.001/50Â Â = 0.00002M

Â Â Â Â Â [H+] = M = 0.00002M

Â Â Â Â PH = -log0.00002

Â Â Â Â Â Â Â Â Â Â Â Â = 4.6989

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