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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