Answer:
![pH=5](https://img.qammunity.org/2019/formulas/chemistry/high-school/9dcxhd9l8mkbxf4o1uto9yegbr542h1at5.png)
Step-by-step explanation:
Hello,
In this case, for the dissociation of the sodium hydroxide into sodium and hydroxyl ions we have:
![NaOH\rightarrow Na^++OH^-](https://img.qammunity.org/2019/formulas/chemistry/high-school/mff7s6zanvz2ctky86enlhyvn0zgaydnrl.png)
Which is a 100% dissociation as sodium hydroxide is a strong base, therefore, the concentration of the hydroxyl ions are 1.0x10⁻⁹. For that reason, the pOH could be computed as:
![pOH=-log([OH^-})=-log(1.0x10^(-9))=9](https://img.qammunity.org/2019/formulas/chemistry/high-school/yg2gtn7vswh5wyxjqy2pcalmzhm9h5ge1u.png)
Finally, from the definition of pH, we have:
![pH+pOH=14\\pH=14-pOH=14-9\\pH=5](https://img.qammunity.org/2019/formulas/chemistry/high-school/yxhs9a2y6vbowm91a9ph2smswvoha12i8r.png)
Best regards.