Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH)2(s) in a solution buffered at a pH of 11.45

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally