Calculate the molarity (M) of NaOH solution which required 22.15 mL of NaOH to r

Question

Calculate the molarity (M) of NaOH solution which required 22.15 mL of NaOH to react completely with 0.552 g KHP. This problem is similar to what is done in the Lab and calculation steps shown (Part A).. Show all steps by using conversion factors. Must finish with c.v. and f.v. Calculate the volume (mL) of 0.450 M of KOH required to neutralize 25.55 mL of 0.155 M HCI solution. Show Equation and rearranged Equation before solving the problem. Show all steps with conversion factor and formula, equations etc. to get full credit. A sample of KHP is contaminated with NaCI is used to standardize NaOH solution and find the molarity of NaOH. How would this titration affect the volume of NaOH needed to neutralize the add KHP. The volume of NaOH would be high or low ?

Explanation / Answer

2) Balanced equation for reaction between KOH and HCl is

KOH + HCl ------------> KCl + H2O

KOH :

Molarity M1 = 0.45 M

Volume V1 = ? mL

moles n1 = 1

HCl:

Molarity M2 = 0.155 M

Volume V2 = 25.55 mL

moles n2 = 1

At equivalence point,

    M1V1/n1 = M2V2/n2

V1 = (M2V2/n2) (n1/M1)

        = (0.155 M X 25.55 mL/ 1) ( 1/ 0.45 M)

          = 8.8 mL

V1 = 8.8 mL

Therefore,

volume of the KOH solution required = 8.8 mL

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