Chemistry, 15.04.2020 00:52 calebcoolbeans6691

A chemist titrates 160.0 mL of a 0.3065 M cyanic acid (HCNO) solution with 0.4994 M NaOH solution at 25 °C. Calculate the pH at equivalence. The pK of cyanic acid is 3.46 Round your answer to 2 decimal places.

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Answer from: andresduenas72

The pH at equivalence for the given solution is 8.37.

Explanation:

We know that,

$pK_{a} = -log K_{a}$

3.46 = $-log K_{a}$

$K_{a} = 3.467 \times 10^{-4}$

Now, we will calculate the volume of NaOH required to reach the equivalence point as follows.

$M_{(HCNO)} \times V_{(HCNO)} = M_{(NaOH)} \times V_{(NaOH)}$

$0.3065 M \times 160.0 mL = 0.4994M \times V_{(NaOH)}$

$V_{(NaOH)}$ = 98.1978 mL

The given data is as follows.

M(HCNO) = 0.3065 M

V(HCNO) = 160 mL

M(NaOH) = 0.4994 M

V(NaOH) = 98.1978 mL

So, moles of HCNO will be calculated as follows.

mol(HCNO) = $M(HCNO) \times V(HCNO)$

mol(HCNO) = $0.3065 M \times 160 mL$

= 49.04 mmol

Now, the moles of NAOh will be calculated as follows.

mol(NaOH) = $M(NaOH) \times V(NaOH)$

mol(NaOH) = $0.4994 M \times 98.1978 mL$

= 49.04 mmol

This means that, 49.04 mmol of both HCNO and NaOH will react to form $CNO^{-}$ and $H_{2}O$ .

Here, $CNO^{-}$ is strong base . So,

$CNO^{-}$ formed = 49.04 mmol

Total volume of the solution is as follows.

Volume of Solution = 160 + 98.1978 = 258.1978 mL

And,

$K_{b}$ of $CNO^{-}$ = $\frac{K_{w}}{K_{a}}$

= $\frac{1 \times 10^{-14}}{3.467 \times 10^{-4}}$

= $2.884 \times 10^{-11}$

The concentration of $CNO^{-}$ is as follows.

c = $\frac{49.04 mmol}{258.1978 mL}$

= 0.1899M

Also,

$CNO^{-} + H_{2}O \rightarrow HCNO + OH^{-}$

Initial: 0.1899 0 0

Equilibm:0.1899 - x x x

$K_{b} = \frac{[HCNO][OH^{-}]}{[CNO^{-}]}$

$K_{b} = \frac{x \times x}{(c - x)}$

We assume that x can be ignored as compared to c . Hence, above formula can be rewritten as follows.

$K_{b} = \frac{x \times x}{c}$

so, x = $\sqrt (K_{b} \times c)$

x = $\sqrt ((2.884 \times 10^{-11}) \times 0.1899)$

= $2.34 \times 10^{-6}$

Here, c is much greater than x, this means that our assumption is correct .

so, x = $2.34 \times 10^{-6}$ M

$[OH^{-}] = x = 2.34 \times 10^{-6}$ M

As,

pOH = $-log [OH^{-}]$

= $-log (2.34 \times 10^{-6})$

= 5.6307

Also, pH = 14 - pOH

= 14 - 5.6307

= 8.3693

or, = 8.37 (approx)

Thus, we can conclude that the pH at equivalence for the given solution is 8.37.

Answer from: Quest

the formula for the conversion of temperature from degree celsius to kelvin is:

$k = ^{o}c + 273.15$

the sample of water at $38^{o}c$ is converted to kelvin temperature by using the above formula as:

$k = 38^{o}c + 273.15 = 311.15 k$

on comparing the two samples of water at $38^{o}c$ or we can say that at $311.15 k$ and at $295 k$ respectively. the difference between the temperatures of two samples of water is:

$311.15 k - 295 k = 16.15 k$

hence, one can say that the sample of water at $38^{o}c$ is greater than the sample of water at $295 k$ by $16.15 k$ .

Answer from: Quest

its pwirhfpiwe

explanation:

because dgjpwojefopjwp

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A chemist titrates 160.0 mL of a 0.3065 M cyanic acid (HCNO) solution with 0.4994 M NaOH solution at...

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