Answer to Question #93173 in General Chemistry for Danielle

Question

In titration 1 of the initial reaction mixture (before H2SO4 is added), 17.5 mL of 0.500 M NaOH is required to titrate a 1.00 mL sample from a mixture of 10.00 mL of CH3COOH (d = 1.05 g/mL) and 10.00 mL of C3H7OH (d = 0.80 g/mL). Show the calculation of the initial [CH3COOH] and [C3H7OH] by using the volume and density data.

Answer 1

The titration will occur according the following equation:

CH₃COOH + NaOH = CH₃COONa + H₂O

The amount of NaOH in moles is:

$n(NaOH)=V(NaOH)*C(NaOH)=17.5*10^{-3}L*0.500M=8.75*10^{-3}mol$

Thus, the amount of acetic acid in moles was the same as solium hydroxide 8.75*10^-3mol.

The alcohol will not be titrated by NaOH and it will probably form a portion of ester. Thus, we will titrate acetic acid only. Let us write down all required starting parameters of the initial mixture:

$M(CH_{3}COOH)=60.05g/mol$

$V(CH_{3}COOH)=10mL$

$d(CH_{3}COOH)=1.05g/mL$

$M(C_{3}H_{7}OH)=60.10g/mol$

$V(C_{3}H_{7}OH)=10mL$

$d(C_{3}H_{7}OH)=0.80g/mL$

Now, we can calculate the amount in moles of both mixture components:

$m(CH_{3}COOH)=d(CH_{3}COOH)*V(CH_{3}COOH)=10mL*1.05g/mL=10.5g$

$n(CH_{3}COOH)=\frac{m(CH_{3}COOH)}{M(CH_{3}COOH)}=\frac{10.5g}{60.05g/mol}=0.175mol$

$m(C_{3}H_{7}OH)=d(C_{3}H_{7}OH)*V(C_{3}H_{7}OH)=0.80g/mL*10mL=8.0g$

$n(C_{3}H_{7}OH)=\frac{m(C_{3}H_{7}OH)}{M(C_{3}H_{7}OH)}=\frac{8.0g}{60.10g/mol}=0.133mol$

Now it is a right time to assume the following approximation about the total volume of the mixture, which is up to some extent must be the simple additive of the volumes of each component:

$V(mixture)=V(CH_{3}COOH)+V(C_{3}H_{7}OH)=10mL+10mL=20mL$

Finally, the initial concentrations of the components are:

$C(CH_{3}COOH)=\frac{n(CH_{3}COOH)}{V(mixture)}=\frac{0.175mol}{0.02L}=8.75M$

$C(C_{3}H_{7}OH)=\frac{n(C_{3}H_{7}OH)}{V(mixture)}=\frac{0.133mol}{0.02L}=6.65M$

It is noticeable that the amount of CH₃COOH is totally the same as it was calculated on the base of the required NaOH amount as a titrant.