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A chemist titrates of a hypochlorous acid solution with solution at . Calculate the pH at equivalence. The of hypochlorous acid

const2013 [10]

The question is incomplete, here is the complete question:

A chemist titrates 110.0 mL of a 0.2412 M hypochlorous acid (HCIO) solution with 0.0613 M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of hypochlorous acid is 7.50. Round your answer to 2 decimal places

Answer: The pH of the solution is 10.09

Explanation:

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HClO$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.2412M\\V_1=110.0mL\\n_2=1\\M_2=0.0613M\\V_2=?mL$

Putting values in above equation, we get:

$1\times 0.2412\times 110.0=1\times 0.0613\times V_2\\\\V_2=\frac{1\times 0.2412\times 110.0}{1\times 0.0613}=432.8mL$

At equivalence, the number of moles of acid is equal to the number of moles of base. Also, the moles of salt which is NaClO will also be the same.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For HClO:

Molarity of HClO solution = 0.2412 M

Volume of solution = 110.0 mL

Putting values in equation 1, we get:

$0.2412M=\frac{\text{Moles of HClO}\times 1000}{110}\\\\\text{Moles of HClO}=\frac{(0.2412\times 110)}{1000}=0.026532mol$

For NaClO:

Moles of NaClO = 0.026532 moles

Volume of solution = [432.8 + 110] mL = 542.8 mL

Putting values in above equation, we get:

$\text{Molarity of NaClO}=\frac{0.026532\times 1000}{542.8}=0.0489M$

To calculate the pH of the solution, we use the equation:

$pH=7+\frac{1}{2}[pK_a+\log C]$

where,

$pK_a$ = negative logarithm of weak acid which is hypochlorous acid = 7.50

C = concentration of the salt = 0.0489 M

Putting values in above equation, we get:

$pH=7+\frac{1}{2}[7.50+\log (0.0489)]\\\\pH=7+3.09=10.09$

Hence, the pH of the solution is 10.09

4 0

2 years ago

An atom

Natasha2012 [34]

Answer:

Doesnt take up space

Explanation:

Looked it up lol

3 0

3 years ago

A sheet of gold weighing 8.8 g and at a temperature of 10.5°C is placed flat on a sheet of iron weighing 19.5 g and at a tempera

timofeeve [1]

Answer:

$T = 36.393\,^{\textdegree}C$

Explanation:

The contact between the sheet of gold and the sheet of iron allows a heat transfer until thermal equilibrium is done, which means that both sheets have the same temperature:

$-Q_{iron} = Q_{gold}$

$-(0.008\,kg)\cdot (452\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-54.4\,^{\textdegree}C) = (0.0195\,kg)\cdot (129\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-10.5\,^{\textdegree}C)$

$-(3.616\,\frac{J}{^{\textdegree}C})\cdot (T-54.4\,^{\textdegree}C) = (2.515\,\frac{J}{^{\textdegree}C})\cdot (T-10.5^{\textdegree}C)$

$-1.438\cdot (T - 54.4^{\textdegree}C) = T-10.5^{\textdegree}C$

$-1.438\cdot T +78.227^{\textdegree}C = T - 10.5^{\textdegree}C$

$2.438\cdot T = 88.727\,^{\textdegree}C$

The final temperature is:

$T = 36.393\,^{\textdegree}C$

8 0

2 years ago

Read 2 more answers

How many milliliters of nitrogen, N2, would have to be collected at 99.19 kPa and 28oC to have a sample containing 0.015 moles o

Semmy [17]

Answer:

378mL

Explanation:

The following data were obtained from the question:

Pressure (P) = 99.19 kPa

Temperature (T) = 28°C

Number of mole (n) = 0.015 mole

Volume (V) =...?

Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:

For Pressure:

101.325 KPa = 1 atm

Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 28°C

T(K) = 28°C + 273 = 301K.

Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:

PV = nRT

Pressure (P) = 0.98 atm

Temperature (T) = 301K

Number of mole (n) = 0.015 mole

Gas constant (R) = 0.0821atm.L/Kmol.

Volume (V) =...?

0.98 x V = 0.015 x 0.0821 x 301

Divide both side by 0.98

V = (0.015 x 0.0821 x 301) /0.98

V = 0.378 L

Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:

1L = 1000mL

Therefore, 0.378L = 0.378 x 1000 = 378mL

Therefore, the volume of N2 collected is 378mL

4 0

2 years ago

Which stament about energy is mostly correct

klasskru [66]

I would say C is the most correct.
In D it depends on what water source you're using. Let's say it is a waterfall, then the source of the water (melting ice or a lake) may disappear in the future.
If you're using underwater "windmills" placed in the ocean, then you would expect it to last a while as the ocean will not disappear in the near future.

6 0

3 years ago