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2 years ago

Write ionic equations for the following: HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

Chemistry

1 answer:

vodomira [7]2 years ago

4 0

Answer:

HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)

Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:

moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl

moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH

You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC

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How much sodium bicarbonate to raise alkalinity in pool.

almond37 [142]

Answer:

A rule of thumb is that 1.5 lbs. of baking soda per 10,000 gallons of water will raise alkalinity by about 10 ppm. If your pool's pH is tested below 7.2, add 3-4 pounds of baking soda. If you're new to adding pool chemicals, start by adding only one-half or three-fourths of the recommended amount.

4 0

2 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

Fed [463]

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$

$4.2 + log\frac{0.05}{0.045} = 4.245$

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

$pH = pKa + log\frac{base}{acid}$

$4.245 = 3.75 + log\frac{base}{acid}$

$log\frac{base}{acid} = 0.5$

$\frac{base}{acid} = 3.162$

Now we must solve the equation above. This will be done using the following values

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L$

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

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3 0

2 years ago

Please help me will mark brainliest

LuckyWell [14K]

Answer:

accretion

Explanation:

the coming together and cohesion of matter under the influence of gravitation to form larger bodies.

8 0

3 years ago

15.00 g of NH4HS(s) is introduced into a 500. mL flask at 25 °C, the flask is sealed, and the system is allowed to reach equilib

Ahat [919]

Answer:

0.328 atm

Explanation:

Kp is the equilibrium constant calculated based on the pressure, and it depends only on the gas substances. It will be the multiplication of partial pressures of the products raised to their coefficients divided by the multiplication of partial pressures of the reactants raised to their coefficients.

For the equation given, the stoichiometry is 1 mol of NH₃ for 1 mol of H₂S, so they will have the same partial pressure in equilibrium, let's call it p. So:

Kp = pxp

0.108 = p²

p = √0.108

p = 0.328 atm, which is the partial pressure of the ammonia.

3 0

3 years ago

How many moles of HNO3 are present if 4.90×10−2 mol of Ba(OH)2 was needed to neutralize the acid solution?

Amiraneli [1.4K]

Answer:

0.098 moles

Explanation:

Let y represent the number of moles present

1 mole of Ba(OH)₂ contains 2 moles of OH- ions.

Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.

To get the y moles, we then do cross multiplication

1 mole * y mole = 2 moles * 0.049 mole

y mole = 2 * 0.049 / 1

y mole = 0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present.

3 0

3 years ago