1. How many mL of 0.200 M potassium hydroxide are required to deliver 0.045 mol.

Question

1. How many mL of 0.200 M potassium hydroxide are required to deliver 0.045 mol.

2. What is the final concentration of a solution prepared by diluting 6.00 mL of a 2.24 M solution of magnesium nitrate to a final volume of 300.0 mL?

3. A 25.0 mL of 0.50 M NaOH is titrated until neutralized into a 50.0 mL sample of HCL. What was the concentration of the HOLY?

4. Titration reveals that 11.6 mL of 3.0 M sulfuric acid are required to neutralize the sodium hydroxide in 25.00 mL of NaOH solution. What is the molar its of the NaOH solution?

Explanation / Answer

1. A c to No of mole sMolarity (M) = No of moles x 1000 / Volume

Volume = No of moles x Molarity /1000

Volume = 0.045 x 0.2 /1000 = 0.9x 10-5 mL

2. A/c to M1V1 =M2V2

M2= M1V1/V2

M2= 2.24 x 6 / 300

M2= 0.044 M

3. A/c to MaVa = MbVb

Ma = MbVb /Va

Ma = 0.5 x 25/ 50

Ma = 0.25 M

4. A/c to MaVa = MbVb

Mb = MaVa / Vb

Mb = 3 x 11.6/25

Mb = 1.392 M

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