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The molarity of the NaOH solution is 0.03 M
We'll begin by calculating the mole of the KHP
- Mass = 0.212 g
- Molar mass = 204.22 g/mol
- Mole of KHP =?
Mole = mass /molar mass
Mole of KHP = 0.212 / 204.22
Mole of KHP = 0.001 mole
Next, we shall determine the molarity of the KHP solution
- Mole of KHP = 0.001 mole
- Volume = 50 mL = 50/1000 = 0.05 L
- Molarity of KHP =?
Molarity = mole / Volume
Molarity of KHP = 0.001 / 0.05
Molarity of KHP = 0.02 M
Finally , we shall determine the molarity of the NaOH solution
KHP + NaOH —> NaPK + H₂O
From the balanced equation above,
- The mole ratio of the acid, KHP (nA) = 1
- The mole ratio of base, NaOH (nB) = 1
From the question given above, the following data were obtained:
- Volume of acid, KHP (Va) = 50 mL
- Molarity of acid, KHP (Ma) = 0.02 M.
- Volume of base, NaOH (Vb) = 35 mL
- Molarity of base, NaOH (Mb) =?
MaVa / MbVb = nA / nB
(0.02 × 50) / (Mb × 35) = 1
1 / (Mb × 35) = 1
Cross multiply
Mb × 35 = 1
Divide both side by 35
Mb = 1 / 35
Mb = 0.03 M
Thus, the molarity of the NaOH solution is 0.03 M
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Answer: c. +4
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here manganese is having an oxidation state of +4 called as cation and oxygen forms an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged to give which written in simplest whole number ratios to give neutral