Consider this reaction:

A student uses a solution of 1.2 molar sodium hydroxide (NaOH) to calculate the concentration of a solution of sulfuric acid (H2

Mila [183]

From the calculations, the concentration of the acid is 0.24 M.

<h3>What is neutralization?</h3>

The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.

We have to use the formula;

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O

CA = ?

CB = 1.2 M

VA = 50 mL

VB = 20 mL

NA = 1

NB = 2

CA = CBVBNA/VANB

CA = 1.2 M * 20 mL * 1/ 50 mL * 2

CA = 0.24 M

Learn more about neutralization:brainly.com/question/27891712

#SPJ1

5 0

2 years ago

Please help due right now <br> 10 POINTS

Darya [45]

Answer:

F=ma

Explanation:

F=m×a

according to that F÷m=a and also F ÷a=m

3 0

3 years ago

What's the chemical equation for the carbon dioxide bubbles in a homemade lava lamp?

julsineya [31]

Answer:

C₆H₈O₇+ 3NaHCO₃ --› Na₃C₆H₅O₇ + 3CO2 + 3H₂O

Explanation:

The reaction occuring in lava lamp is acid base reaction.

When you drop tablet into water the citric acid reacts with sodium bicarbonate and forms water, a salt, and bubbles of carbon dioxide gas.

6 0

3 years ago

Read 2 more answers

I need help with this

son4ous [18]

Answer:

2nd one

Explanation:

4 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago