Question

Calculate the volume of a 0.500M NaOH solution needed to neutralize (titrate to the endpoint):

a)10.0 mL 0.300M HCl


b)10.0 mL 0.200M H2SO4


c) 4.08 g KHP (molar mass 204 g/mol

Answer 1

Answer on Question #48803, Chemistry, Other

Task:

Calculate the volume of a 0.500 M NaOH solution needed to neutralize (titrate to the endpoint):

a) 10.0 mL 0.300M HCl

b) 10.0 mL 0.200M H₂SO₄

c) 4.08 g KHP (molar mass 204 g/mol)

Answer:

$$\mathrm{NaOH} + \mathrm{HCl} = \mathrm{NaCl} + \mathrm{H_2O}$$ $$C_M(\mathrm{NaOH}) \cdot V(\mathrm{NaOH}) = C_M(\mathrm{HCl}) \cdot V(\mathrm{HCl})$$ $$V(\mathrm{NaOH}) = \frac{0,300 \cdot 10,0}{0,500} = 6,00 \text{ ml}$$ $$2\mathrm{NaOH} + \mathrm{H_2SO_4} = \mathrm{Na_2SO_4} + 2\mathrm{H_2O}$$ $$2\nu(\mathrm{NaOH}) = \nu(\mathrm{Na_2SO_4})$$ $$\nu(\mathrm{NaOH}) = \frac{\nu(\mathrm{Na_2SO_4})}{2}$$ $$C_M(\mathrm{NaOH}) \cdot V(\mathrm{NaOH}) = \frac{C_M(\mathrm{Na_2SO_4}) \cdot V(\mathrm{Na_2SO_4})}{2}$$ $$V(\mathrm{NaOH}) = \frac{0,200 \cdot 10,0}{2 \cdot 0,500} = 2 \text{ ml}$$ $$\mathrm{NaOH} + \mathrm{KHP} = \mathrm{KNaP} + \mathrm{H_2O}$$ $$\nu(\mathrm{NaOH}) = \frac{m(\mathrm{NaOH})}{M((\mathrm{NaOH}))}$$ $$\nu(\mathrm{NaOH}) = \nu(\mathrm{KNaP})$$ $$\nu(\mathrm{NaOH}) = \frac{4,08}{204} = 0,02 \text{ mol}$$ $$C_M(\mathrm{NaOH}) \cdot V(\mathrm{NaOH}) = \nu(\mathrm{NaOH})$$ $$V(\mathrm{NaOH}) = \frac{\nu(\mathrm{NaOH})}{C_M(\mathrm{NaOH})} = \frac{0,02}{0,500} = 0,04 \text{ ml}$$

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