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10 days ago

28 ml of 0.10 m hcl is added to 60 ml of 0.10 m sr(oh)2. determine the concentration of oh− in the resulting solution.

2 answers:

5 0

Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]

The ions H+ and OH- neutralize each other. If the amounts are not equal, one ion will be in excess.
Follow these steps:[

1. Determine the moles of ions: mole= Molarity * Volume (in liters); n= M * V OR millimoles = Molarity * Volume (in milliliters);2. Write the balanced chemical equation
3. Identify the excess ion4. Use the total volume (Volume of acid + Volume of base) to calculate the concentration of the excess ion.

n HCI = 28 ml * 0.10 M = 0.28 mmol, which releases 0.28 mmol of H+ ions
n Sr(OH)2= 60 ml * 0.10 M = 0.60 mmol, releases 2*0.60=1.20 mmol of OH- ions as Sr(OH)2⇒ Sr2+ + 2OH-
The neutralization reaction can be expressed as OH- + H+ ---> H2O. Since the ratio is 1:1, it implies that 1 mmol of hydroxide ions neutralizes 1 mmol of hydrogen ions. Given that there are more OH- ions, they will be in excess

n(OH-) - n(H+) = 1.20 - 0.28 = 0.92 mmol of OH- ions remaining unreacted.

Total volume = Volume of acid + Volume of base = 28 ml + 60 ml = 98 ml

Concentration of OH- ions = moles / total volume = 0.92/98 = 0.009 M

Thus, the resulting concentration is 0.009 M.

alisha [2.7K]10 days ago

5 0

Answer: 0.52 M

To determine the moles for the specified molarity, the following equation is employed:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

The molarity of $HCl$ solution = 0.10 M

Volume of $HCl$ solution = 28 mL = 0.028 L

Substituting values into equation 1, we obtain:

$\text{Moles of} HCl={0.10}\times{0.028}=0.0028moles$

Molarity of $Sr(OH)_2$ solution = 0.10 M

Volume of $Sr(OH)_2$ solution = 60 mL = 0.060 L

Substituting values into equation 1, we gather:

$\text{Moles of} Sr(OH)_2={0.10}\times{0.060}=0.006moles$

$2HCl+Sr(OH)_2\rightarrow SrCl_2+2H_2O$

In accordance with stoichiometry:

2 moles of $HCl$ neutralize 1 mole of $Sr(OH)_2$

Thus, 0.0028 moles of $HCl$ neutralize= $\frac{1}{2}\times 0.0028=0.0014moles$ of $Sr(OH)_2$

Consequently, (0.006-0.0014) moles = 0.046 moles of $Sr(OH)_2$ remain in 88 ml of solution.

The concentration of $[OH^-]$ can be calculated as $\frac{moles}{\text {Volume in L}}=\frac{0.046}{0.088}=0.52M$

Thus, the concentration of $[OH^-]$ in the final solution is 0.52 M

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1 month ago

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27 days ago

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lorasvet [2589]

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To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.

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Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:

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Mass of N2 = 0.346 × 28

Mass of N2 = 9.69g

Hence, 7.744L of N2 consists of 9.69g of N2

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