Question

there is a problem in which they ask me the pH of a reaction between NaOH 50mL and 47mL of HCL of 0,5M.The result it says its 7, but i don´t know how to solve it?.Please any help.Thanks

Answer 1

Answer on Question #58643, Chemistry / General Chemistry

There is a problem in which they ask me the pH of a reaction between NaOH 50mL and 47mL of HCL of 0,5M. The result it says its 7, but i don't know how to solve it?. Please any help. Thanks.

Solution:

After addition of 47 ml 0.5 M NSL solutions to 50 ml 0,5M of NaOH solution the total amount of solution is equal $50 + 47 = 97$ ml.

Concentration of alkali is equal in solution:

$$[ \mathrm {O H} ^ {-} ] = \frac {5 0 \cdot 0 . 5 - 4 7 \cdot 0 . 5}{9 7} = 0, 0 1 5 4 6 M$$ $$\mathrm {p O H} = - \lg [ \mathrm {O H} ^ {-} ] = - \lg (0. 0 1 5 4 6) = 1. 8 1$$ $$p H = 1 4 - p O H - 1 4 - 1. 8 1 = 1 2. 1 9$$

In order that pH solution it was equal 7, concentration of the added HCl has to be a little higher, than concentration of initial NaOH solution:

$$V _ {N a O H} \cdot M _ {N a O H} = V _ {H C l} \cdot M _ {H C l}$$ $$M _ {N a O H} = V _ {H C l} \cdot M _ {H C l} / V _ {N a O H} = 4 7 \cdot 0. 5 / 5 0 = 0. 4 7 \mathrm {m o l} / \mathrm {l}$$

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