1. ## Calculating ion concentrations

Calculate the concentration of hydronium ions in each solution.
-->17.9mL of 0.175mol/L HNO3 added to 35.4mL of 0.0160mol/L Ca(OH)2

Calculate the concentration of hydroxide ions in each solution.
-->16.5mL of 1.50mol/L H2SO4 added to 12.7mL of 5.50mol/L NaOH.

Thanks!

2. Calculate the concentration of hydronium ions in each solution.
-->17.9mL of 0.175mol/L HNO3 added to 35.4mL of 0.0160mol/L Ca(OH)2
$
\frac {0.0179*0.175*1-0.0354*0.0160*2}{.0179+.0354}=.0375M$

All you are doing is multiplying the number of Litres by the number of molecules in each Litre to get the number of molecules then multiplying by the number of ions in a molecule to get the number of ions. Then you cancel out the Hydronium and Hydroxide ions that combine to form water and divide by the number of Litres to get the number of ions per Litre. Have a shot at the second one yourself.

3. this is what I got for the second one...

H2SO4=0.0165L x 1.50mol/L NaOH=0.0354L x 0.0160mol/L
= 0.02475mol = 0.0005664mol

Amount excess: 0.02475-0.0005664
= 0.0241836mol

Total Volume=0.0165L+0.0005664L
=0.0170664L

A: 0.02418mol/0.01707L
= 1.42mol/L

...but the answer is suppose to be 0.697mol/L

4. There are 2 mistakes here, but I think have got the general idea. First, $H_2SO_4$ will donate 2 $H^+$ ions so you need to multiply the number of mol of $H_2SO_4$ by 2 to get the number of $H^+$ ions.

Total Volume=0.0165L+0.0005664L
=0.0170664L
You have used the number of mol instead of the volume but I think you know what you are doing here.