Write down the NAME or FORMULA of the substance that acts as an ampholyte in the above equations. Give a…

Question

Write down the NAME or FORMULA of the substance that acts as an ampholyte in the above equations. Give a reason for the answer.

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to your questions:

8.1.1 Define a base in terms of the Lowry-Brønsted theory.
A base is a substance that acts as a proton acceptor.

8.1.2 Write down the NAME or FORMULA of the substance that acts as an ampholyte in the above equations. Give a reason for the answer.
Substance: $\boxed{\text{HSO}_4^-}$ (Hydrogen sulfate ion)
Reason: In equation II, $\text{HSO}_4^-$ donates a proton to $\text{H}_2\text{O}$ (acting as an acid). In the reverse of equation I, $\text{HSO}_4^-$ accepts a proton from $\text{H}_3\text{O}^+$ to form $\text{H}_2\text{SO}_4$ (acting as a base).

8.2.1 Calculate the pH of this nitric acid solution.
Given: Concentration of $\text{HNO}_3 = 0.096 \text{ mol}\cdot\text{dm}^{-3}$ .
$\text{HNO}_3$ is a strong acid, so it dissociates completely:

\text{HNO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{O}^+(\text{aq}) + \text{NO}_3^-(\text{aq})

Determine the concentration of hydronium ions.
Since $\text{HNO}_3$ is a strong acid, $[\text{H}_3\text{O}^+] = [\text{HNO}_3]$ .
$[\text{H}_3\text{O}^+] = 0.096 \text{ mol}\cdot\text{dm}^{-3}$
Calculate the pH.
$\text{pH} = -\log[\text{H}_3\text{O}^+]$ $\text{pH} = -\log(0.096)$ $\text{pH} \approx \boxed{1.02}$
8.2.2 Calculate the concentration of the EXCESS $\text{HNO}_3$ in the new solution.
Balanced equation: $2\text{HNO}_3(\text{aq}) + \text{Na}_2\text{CO}_3(\text{aq}) \rightarrow 2\text{NaNO}_3(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})$
Given:
Volume of $\text{HNO}_3$ ( $V_A$ ) = $50 \text{ cm}^3 = 0.050 \text{ dm}^3$
Concentration of $\text{HNO}_3$ ( $C_A$ ) = $0.096 \text{ mol}\cdot\text{dm}^{-3}$
Volume of $\text{Na}_2\text{CO}_3$ ( $V_B$ ) = $25 \text{ cm}^3 = 0.025 \text{ dm}^3$
Concentration of $\text{Na}_2\text{CO}_3$ ( $C_B$ ) = $0.075 \text{ mol}\cdot\text{dm}^{-3}$
Calculate the initial moles of $\text{HNO}_3$ .
$n(\text{HNO}_3)_{\text{initial}} = C_A \times V_A$ $n(\text{HNO}_3)_{\text{initial}} = (0.096 \text{ mol}\cdot\text{dm}^{-3}) \times (0.050 \text{ dm}^3)$ $n(\text{HNO}_3)_{\text{initial}} = 0.0048 \text{ mol}$
Calculate the initial moles of $\text{Na}_2\text{CO}_3$ .
$n(\text{Na}_2\text{CO}_3)_{\text{initial}} = C_B \times V_B$ $n(\text{Na}_2\text{CO}_3)_{\text{initial}} = (0.075 \text{ mol}\cdot\text{dm}^{-3}) \times (0.025 \text{ dm}^3)$ $n(\text{Na}_2\text{CO}_3)_{\text{initial}} = 0.001875 \text{ mol}$
Determine the moles of $\text{HNO}_3$ that react with $\text{Na}_2\text{CO}_3$ .
From the balanced equation, the mole ratio $\text{HNO}_3 : \text{Na}_2\text{CO}_3$ is $2:1$ .
$n(\text{HNO}_3)_{\text{reacted}} = 2 \times n(\text{Na}_2\text{CO}_3)_{\text{initial}}$ $n(\text{HNO}_3)_{\text{reacted}} = 2 \times 0.001875 \text{ mol}$ $n(\text{HNO}_3)_{\text{reacted}} = 0.00375 \text{ mol}$
Since $n(\text{HNO}_3)_{\text{initial}} (0.0048 \text{ mol}) > n(\text{HNO}_3)_{\text{reacted}} (0.00375 \text{ mol})$ , $\text{HNO}_3$ is in excess.
Calculate the moles of excess $\text{HNO}_3$ .
$n(\text{HNO}_3)_{\text{excess}} = n(\text{HNO}_3)_{\text{initial}} - n(\text{HNO}_3)_{\text{reacted}}$ $n(\text{HNO}_3)_{\text{excess}} = 0.0048 \text{ mol} - 0.00375 \text{ mol}$ $n(\text{HNO}_3)_{\text{excess}} = 0.00105 \text{ mol}$
Calculate the total volume of the new solution.
$V_{\text{total}} = V_A + V_B$ $V_{\text{total}} = 0.050 \text{ dm}^3 + 0.025 \text{ dm}^3$ $V_{\text{total}} = 0.075 \text{ dm}^3$
Calculate the concentration of the excess $\text{HNO}_3$ .
$C(\text{HNO}_3)_{\text{excess}} = \frac{n(\text{HNO}_3)_{\text{excess}}}{V_{\text{total}}}$ $C(\text{HNO}_3)_{\text{excess}} = \frac{0.00105 \text{ mol}}{0.075 \text{ dm}^3}$ $C(\text{HNO}_3)_{\text{excess}} = \boxed{\text{0.014 mol}\cdot\text{dm}^{-3}}$

Write down the NAME or FORMULA of the substance that acts as an ampholyte in the above equations. Give a reason for the answer.

Step-by-step solution

Determine the concentration of hydronium ions.

Calculate the pH.

Calculate the initial moles of $\text{HNO}_3$ .

Calculate the initial moles of $\text{Na}_2\text{CO}_3$ .

Determine the moles of $\text{HNO}_3$ that react with $\text{Na}_2\text{CO}_3$ .

Calculate the moles of excess $\text{HNO}_3$ .

Calculate the total volume of the new solution.

Calculate the concentration of the excess $\text{HNO}_3$ .

Still stuck on this one?

More Chemistry Questions

Write down the NAME or FORMULA of the substance that acts as an ampholyte in the above equations. Give a reason for the answer.

Step-by-step solution

Determine the concentration of hydronium ions.

Calculate the pH.

Calculate the initial moles of HNO3\text{HNO}_3HNO3​.

Calculate the initial moles of Na2CO3\text{Na}_2\text{CO}_3Na2​CO3​.

Determine the moles of HNO3\text{HNO}_3HNO3​ that react with Na2CO3\text{Na}_2\text{CO}_3Na2​CO3​.

Calculate the moles of excess HNO3\text{HNO}_3HNO3​.

Calculate the total volume of the new solution.

Calculate the concentration of the excess HNO3\text{HNO}_3HNO3​.

Still stuck on this one?

More Chemistry Questions

Calculate the initial moles of $\text{HNO}_3$ .

Calculate the initial moles of $\text{Na}_2\text{CO}_3$ .

Determine the moles of $\text{HNO}_3$ that react with $\text{Na}_2\text{CO}_3$ .

Calculate the moles of excess $\text{HNO}_3$ .

Calculate the concentration of the excess $\text{HNO}_3$ .