True or false: scientific theories and models cannot be modified.

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa

NaOH + CH3COOH → CH3COONa + H2O

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M

CH3COONa = 0.0076 / 0.071 = 0.1070M

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.

pH using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

pH = 4.74 + log ( 0.1070/0.1493)

pH = 4.74 + log 0.717

pH = 4.74 + (-0.14)

pH = 4.60.

7 0

3 years ago

The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .