Question

What volume of 80.0 mM potassium maleate buffer, pH 6.5, would be needed to make 0.200 liters of 5.00 mM maleate buffer containing 2.30 M ammonium nitrate? (6 pts.) Submit Answer Tries 0/2 What mass of ammonium nitrate (F.W. 80.04) would be needed? (6 pts.) Submit Answer Tries /2 In the above example, how many -fold has the original buffer been diluted? (6 pts.) Submit Answer Tries 0/2 Should the ammonium nitrate be dissolved in the buffer before or after the buffer has been brought to 0.200 L? (6 pts.) Submit Answer Tries 0/1

Answer 1

1) M₁V₁=M₂V

M_1 V_1 = M_2 V 80 times X = 5 times 0.2 therefore X = 5 times 0.2/80 = 0.0125 L Weight = NEV/1000 = 2.3 times 0.2 times 80.04/1000 = 0.0368184g Always the solid is to be dissolved and diluted which is the correct format of diluting because in the opposite way the solids may not get dissolved and may occur a change in concentration.

2.)

  

  

4.) Always the solid is to be dissolved and diluted which is the correct format of diluting because in the opposite way the solids may not get dissolved and may occur a change in concentration.