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2 years ago

What’s the total ionic equation for NaOH reacting with HCl

Chemistry

1 answer:

exis [7]2 years ago

6 0

Answer:

Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

General Formulas and Concepts:

Atomic Structure

Compounds

Aqueous Solutions

Solubility Rules
States of matter

Stoichiometry

Reaction RxN Prediction
Balancing Reactions RxN

Explanation:

Step 1: Define

NaOH reacting w/ HCl

NaOH is soluble

HCl is soluble

[RxN] NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)

Step 2: Total Ionic Equation

Break up soluble compounds into ionic form.

[T.I.E] Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

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3 years ago

Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base

lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = $6.20*10^{-5}$

So:

$pKa = - log\:(Ka)$

$pKa = - log\:(6.20*10^{-5})$

$pKa = 5 - log\:6.20$

$pKa = 5 - 0.79$

$\boxed{pKa = 4.21}$

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

$pH = pKa + log\:\dfrac{[salt]}{[acid]}$

$pH = 4.21 + log\:\dfrac{0.130}{0.235}$

$pH = 4.21 + log\:0.55$

$pH = 4.21 + (-0.26)$

$pH = 4.21 - 0.26$

$\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark$

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

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2 years ago