How much ice at 0°C [Cp = 37.1 J/(mol ·°C), ΔH°fus = 6.01 kJ/mol] must be added to return the solution temperature to [ . . . ]?

I got a mass of #"11.91 g"# of ice.

[Don't fall into the trap of using the same #barC_P# of water for the solution; the solution's specific heat capacity changed after mixing. You'd get #"12.63 g"# if you did that AND accidentally used the #barC_P# of ice for heating the ice AFTER it melted.

Speaking of which, do NOT use the heat capacity of ice... you'd get #"13.56 g"# if you did that but used the correct amount of heat transferred.]

DISCLAIMER: REALLY LONG ANSWER!

First, we should determine the amount of heat transferred into the water, and thus its final temperature.

#11.0 cancel"mL" xx "1 L"/(1000 cancel"mL") xx "12.0 mol HNO"_3/cancel"L" = "0.132 mols HNO"_3#

The heat imparted is then:

#0.132 cancel"mols" xx "33.3 kJ"/cancel("mol HNO"_3) = "4.396 kJ" = "4396 J"#

This heat, assuming conservation of energy, goes into heating #"100.00 mL"# of water. If the water was added to the acid, then it would be heated by the acid dissolution using its own heat capacity.

Using a density of #"0.9970749 g/mL"# for water at #25^@ "C"#, and assuming a constant density (and step-function heat capacity, apparently) through the temperature change,

#q = n_"water"barC_(P,"water")DeltaT_"water"#

#"4396 J" = 100.00 cancel"mL" xx (0.9970749 cancel"g")/cancel"mL" xx cancel"1 mol"/(18.015 cancel"g") cdot "75.3 J/"cancel"mol"cdot""^@ "C" cdot (T_f - 25^@ "C")#

Solving this, the final temperature after the heating becomes:

#T_f = 35.55^@ "C"#

Now we want to return the solution to #25^@ "C"# by adding ice to it... We want the ice to melt and then heat up to #25^@ "C"#, because we are shooting for thermal equilibrium with the ice.

However, the solution has a higher heat capacity, of #"80.8 J/mol"cdot""^@ "C"#, and a new volume of #"111.00 mL"# (assuming additivity), with a new density of #"0.9938447 g/mL"#.

So, more heat than #"4396 J"# would be needed. The heat transfer follows conservation of energy as before:

#q_"soln" + q_"ice" = 0#

#overbrace(n_"soln"barC_(P,"soln")DeltaT_"soln")^"solution" + overbrace(q_1 + q_2)^"ice" = 0#

The heat that must be transferred out of the solution is:

#|q_"soln"| = 111.00 cancel"mL" xx (0.9938447 cancel"g")/cancel"mL" xx cancel"1 mol"/(18.015 cancel"g") cdot "80.8 J/"cancel"mol"cdotcancel(""^@ "C") cdot (25cancel(""^@ "C") - 35.55cancel(""^@ "C"))#

#= |-"5220 J"| = "5220 J"#

We want the total heat coming into the ice to be set up as follows:

#q_"ice" = q_1 + q_2#

#= overbrace(n_"ice"DeltabarH_(fus))^"melting" + overbrace(n_"ice"barC_(P,water)DeltaT_(water))^"heating melted ice"#

#= m_("ice")/"18.015 g/mol" cdot DeltabarH_(fus) + m_("ice")/"18.015 g/mol" cdot barC_(P,water)DeltaT_(water)#

#= m_("ice")/"18.015 g/mol"[DeltabarH_(fus) + barC_(P,water)DeltaT_(water)]#

We should be aware NOT to use #barC_P# of the ice... it is not ice after melting. From the conservation of energy equation,

#-q_"soln" = q_"ice"#

#=> "5220 J" = m_("ice")/"18.015 g/mol"[DeltabarH_(fus) + barC_(P,water)DeltaT_(water)]#

Evaluating the bracketed term first, we get:

#DeltabarH_(fus) + barC_(P,water)DeltaT_(water)#

#= (6.01 cancel"kJ")/"mol" xx ("1000 J")/cancel"1 kJ" + "75.3 J"//"mol"cdotcancel(""^@ "C") cdot (25cancel(""^@ "C") - 0cancel(""^@ "C"))#

#=# #"7892.5 J/mol"#

As a result, the mass of ice must be:

#color(blue)(m_("ice")) = (5220 cancel"J")/(7892.5 cancel"J"//cancel"mol") xx "18.015 g"/cancel"mol"#

#=# #color(blue)("11.91 g ice")#