Chemistry pH problem help NaOH and HCl

[marcuss]

Chemistry pH problem help please NaOH and HCl

Calculate the pH after the following total volumes of .32M HCl have been added to 25 mL of 0.16M NaOH
a. 0 mL c. 12.4 mL e. 12.6 mL
b. 1 mL d. 12.5 mL f. 15.0 mL



A. i have answered correctly but i can't seem to get the right answer for any of the others i tried putting them in mmol but it never seems to work if someone could just tell me how to do b through f i would be very pleased thank you.

 

[symbolipoint]

The exercise is a neutralization reaction. Determine which moles of NaOH or HCl is in excess. That would be the one which affects the pH. A relationship which is helpful is 14 = pH + pOH

 

[Blueshift5]

Hello,

Thank you for reaching out for help with this chemistry pH problem. It seems like you have successfully calculated the pH for the first scenario where 0 mL of HCl is added. To calculate the pH for the other scenarios, you will need to use the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is pH = pKa + log([base]/[acid]), where pKa is the negative logarithm of the acid dissociation constant, [base] is the concentration of the base (in this case NaOH), and [acid] is the concentration of the acid (in this case HCl).

To solve for the pH in these scenarios, you will first need to calculate the concentration of HCl after the given volume has been added. To do this, you can use the dilution formula: M1V1 = M2V2, where M1 and V1 are the initial concentration and volume of HCl, and M2 and V2 are the final concentration and volume after the addition of HCl.

Once you have calculated the concentration of HCl, you can plug it into the Henderson-Hasselbalch equation along with the concentration of NaOH (0.16M) to solve for the pH. Remember to convert the volumes to liters before using the equation.

I hope this helps you to successfully calculate the pH for scenarios b through f. If you are still having trouble, please provide your calculations and I will be happy to review them and offer further assistance.

Best of luck with your problem-solving!