Can Someone Help Me Solve The Two pH Problems PLEASE IM LOST?

Question

I Have No Idea How To Solve Them, There Is Not Any Info In My Text

5.54 mL of 0.0557 M HCl is added to 3.37 mL of 0.0377 M NaOH. The pH of the resultant solution is______

5.42 mL of an HCl solution whose pH is 1.84 is added to 3.21 mL of an NaOH solution whose pH was 9.39. The pH of the resultant solution is _____

Answer 1

HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O

What you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.

For your first problem:

0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL

0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH

3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL

0.00542L + 0.00337L = .00879L

[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL

pH = -log(.02) = 1.70

Answer 2

For the first one Figure out the concentration of H+ ( H3O) and then you take the negative log of the value.

For the second convert each Ph to a concentration value using the antilog of the number then find the end concentration and take the negative log of that value