Question

1. write the complete, balanced equation for the reaction of hno3 and naoh.
2. in a titration of hno3 with naoh, the naoh titrant is standardized using khp according to the reaction: khp (aq) + naoh (aq) → h2o (l) + nakp (aq)

a 0.7532 g sample of dry khp is weighed on an analytical balance, placed in a flask, and mixed with 50 ml of water. the initial volume reading on the buret filled with naoh is 0.06 ml. at the endpoint of the titration, the buret reads 32.13 ml. what is the concentration of the naoh? (khp is 204.22 g/mol)

Explanation:

Step1: Calculate moles of KHP

Use the formula $n=\frac{m}{M}$, where $m = 0.7532\ g$ and $M=204.22\ g/mol$.
$n_{KHP}=\frac{0.7532\ g}{204.22\ g/mol}=0.003688\ mol$

Step2: Determine moles of NaOH

From the balanced - chemical equation $KHP(aq)+NaOH(aq)
ightarrow H_2O(l)+NaKP(aq)$, the mole ratio of $KHP$ to $NaOH$ is 1:1. So, $n_{NaOH}=n_{KHP}=0.003688\ mol$

Step3: Calculate volume of NaOH used

The initial volume of $NaOH$ in the buret is $0.06\ mL$ and the final volume is $32.13\ mL$. The volume of $NaOH$ used, $V = 32.13\ mL - 0.06\ mL=32.07\ mL = 0.03207\ L$

Step4: Calculate molarity of NaOH

Use the formula $M=\frac{n}{V}$, where $n = 0.003688\ mol$ and $V = 0.03207\ L$.
$M_{NaOH}=\frac{0.003688\ mol}{0.03207\ L}=0.115\ M$

Step5: Balanced equation for HNO3 and NaOH

The balanced - chemical equation for the reaction of $HNO_3$ and $NaOH$ is $HNO_3(aq)+NaOH(aq)
ightarrow NaNO_3(aq)+H_2O(l)$

Answer:

1. $HNO_3(aq)+NaOH(aq)

ightarrow NaNO_3(aq)+H_2O(l)$

1. $0.115\ M$