STOGS Y11 Titration calculations

How many cm³ in 1 dm³?

• 1000

• 1

• 10

• 0.001

Tags

cm3 to dm3 conversion

How many dm³ in 25cm³?

• 2.5 x 10^-2

• 2.5 x 10^-4

• 2.5 x 10³

• 2.5 x 10^-3

Tags

cm3 to dm3 conversion

How many moles of Na⁺ ions are there in

25cm³ of 0.01 mol/ dm³ NaOH solution?

• 0.25 mol

• 2.5 mol

• 2.5 x 10^-4 mol

• 2.5 x 10^-3 mol

Tags

moles = conc x vol

How many moles of Na⁺ ions are there

in 0.025 dm³ of 1.0 mol / dm³ of Na₂SO₄ solution?

• 0.025 mol

• 2.5 x 10³ mol

• 0.05 mol

• 1.0 x 10^-3 mol

Tags

moles = conc x vol

Numbers of ions released in solution

HCl + NaOH --> NaCl + H₂O

How many moles of NaOH will neutralise 18.3cm³ of 1.0 mol / dm³ HCl solution?

• 6.02 x 10²³ mol

• 1.83 x 10⁴ mol

• 0.0183 mol

• 18.3 mol

Tags

moles = conc x vol

H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

How many moles of H₂SO₄ will neutralise 14.2 cm³ of 0.01 mol/dm³ NaOH? 

• 1.42 x 10⁴ mol

• 7.1 x 10^-5 mol

• 1.42 x 10^-4 mol

• 7.1 x 10⁵ mol

Answer explanation

Each mole of H₂SO₄ neutralises 2 moles of NaOH

Tags

moles = conc x vol

reacting ratio

HCl + NaOH --> NaCl + H₂O

What volume of 0.01 mol/dm³ HCl solution will neutralise

50cm³ of 0.01 mol/dm³ NaOH solution?

• 0.25cm³

• 25cm³

• 50cm³

• 100cm³

Tags

moles = conc x vol

vol = moles / conc

HCl + NaOH --> NaCl + H₂O

What volume of 0.5 mol/dm³ HCl solution will neutralise

25cm³ of 1.0 mol/dm³ NaOH solution?

• 0.5cm³

• 12.5cm³

• 25cm³

• 50cm³

Tags

moles = conc x vol

vol = moles / conc

H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

To 2 dp, what volume of 2.0 mol/dm³ H₂SO₄ solution will neutralise

36.20cm³ of 2.0 mol/dm³ NaOH solution?

• 36.20cm³

• 18.10cm³

• 0.04cm³

• 0.02cm³

Answer explanation

Each mole of H₂SO₄ neutralises 2 moles of NaOH

Tags

moles = conc x vol

vol = moles / conc

reacting ratio

H₃PO₄ + 3NaOH --> Na₃PO₄ + 3H₂O

What volume of 1.0 mol/dm³ H₃PO₄ will neutralise

36.0cm³ of 2.0 mol/dm³ NaOH?

• 216.0cm³

• 36.0cm³

• 72.0cm³

• 24.0cm³

Answer explanation

Each mole of H₃PO₄ neutralises 3 moles of NaOH

Tags

vol = moles / conc

acids as monoprotic/diprotic/triprotic

moles = conc x vol