Вопрос
No 17 1. How many milliliters of 0.10 m H_(2)SO_(4) must be added to 100 mL of 0.10 M NaOH to give a solution that is 0.050 M in H_(2)SO_(4) ? Assume volumes are additive. A samnle of nure KHC_(2)O_(4)cdot H_(2)C_(2)O_(4)cdot 2H_(2)O
Решения
4.1
(294 Голоса)
Римма
элита · Репетитор 8 лет
Ответ
To solve this problem, we need to find the volume of 0.10 M
that needs to be added to 100 mL of 0.10 M NaOH to achieve a final concentration of 0.050 M in
.Given information:- Initial concentration of
: 0.10 M- Initial volume of NaOH solution: 100 mL- Final concentration of
: 0.050 MLet's denote the volume of 0.10 M
to be added as V mL.The total volume of the final solution will be (100 + V) mL.The amount of
in the final solution can be calculated using the formula:Amount of
= Concentration × VolumeThe amount of
in the initial 100 mL of 0.10 M NaOH solution is:Amount of
in NaOH solution = 0.10 M × 0.100 L = 0.010 molThe amount of
in the final solution should be 0.050 M × (100 + V) mL = 0.050 M × (0.100 + V/1000) L = 0.050 × (0.100 + V/1000) molSince the amount of
in the final solution is the sum of the amount of
in the initial NaOH solution and the amount of
added, we can set up the equation:0.010 mol + 0.10 M × V/1000 L = 0.050 × (0.100 + V/1000) molSolving this equation for V, we get:V = 100 mLTherefore, 100 mL of 0.10 M
must be added to 100 mL of 0.10 M NaOH to give a solution that is 0.050 M in
.