Answer to Question #100481 in General Chemistry for Malathy

Question

A. Standardaization of Sodium Hydroxide (NaOH) solution.
TRIAL 1
a)Mass of bottle +Khp = 98.108g
b)Mass of bottle = 97.526g
c)Mass of KHP used = 0.5819g
d)Final buret reading = 32.4ml
e)Initial buret reading = 0ml
f)mL of NaOH used =______?
g)Molarity of NaOH =______?

TRIAL2
a)Mass of bottle +Khp = 109.097g
b)Mass of bottle = 108.51g
c)Mass of KHP used = 0.5868g
d)Final buret reading = 29.4ml
e)Initial buret reading = 0ml
f)mL of NaOH used =______?
g)Molarity of NaOH =______?

TRIAL3
a)Mass of bottle +Khp = 112.442g
b)Mass of bottle = 111.89g
c)Mass of KHP used = 0.5523g
d)Final buret reading = 24.3ml
e)Initial buret reading = 0ml
f)mL of NaOH used =______?
g)Molarity of NaOH =______?

●Calculate Average Molarity =_________?
●Standard deviation=_________?

Answer 1

$$n(KHP) = n (NaOH)$$

$$n(KHP) = m/M$$

$$c(NaOH) = n(NaOH)/V = n(KHP)/V$$

TRIAL1

$$n(KHP) = 0.5819/204.22 = 0.00285 mol$$

Used volume NaOH = 32.4 mL = 0.0324 L

$$c(NaOH) = 0.00285 mol/0.0324L = 0.088M$$

TRIAL2

$$n(KHP) = 0.00287mol$$

Used volume NaOH = 29.4 mL = 0.0294 L

$$c(NaOH) = 0.098 M$$

TRIAL3

$$n(KHP) = 0.00270mol$$

Used volume NaOH = 24.3 mL = 0.0243 L 

$$c(NaOH) = 0.111M$$

Calculate Average Molarity = 0.099 M

Standard deviation= 0.0094